question

Which of the following statements best describes the characteristic of the restoring force in the spring-mass system displaying simple harmonic motion?
a. The restoring force is constant.
b. The restoring force is directly proportional to the displacement of the block.
c. The restoring force is proportional to the mass of the block.
d. The restoring force is maximum when the block is in the equilibrium position.

answer

b (The restoring force is directly proportional to the displacement of the block.)

question

A coordinate system with the origin at the equilibrium position is chosen so that the x coordinate represents the displacement from the equilibrium position. (The positive direction is to the right.)
What is the initial acceleration of the block, a_0, when the block is released at a distance to the right from its equilibrium position?
Express your answer in terms of some or all of the variables A, m, and k.

answer

1. Use hooke's law and newtons second law
F = -kx = ma
2. Solve for a
a = -(kA)/m

question

a coordinate system with the origin at the equilibrium position
is chosen so that the x coordinate represents the displacement from the equilibrium
position. (The positive direction is to the right.)
What is the acceleration a_1 of the block when it passes through its equilibrium position?
Express your answer in terms of some or all of the variables A, m, and k.

answer

Since this is SHM, the force = 0 at equilibrium, x=0. Therefore, if x = 0
F = -kx = ma => a = -(kx)/m => a = 0/m => a=0
Alternatively, if F = 0
F = ma => a = F/m => a = 0/m => a=0

question

In SHM, the magnitude of the block's acceleration reaches its maximum value when the block is ...
a. in the equilibrium position.
b. at either its rightmost or leftmost position.
c. between its rightmost position and the equilibrium position.
d. between its leftmost position and the equilibrium position.

answer

b. (at either its rightmost or leftmost position.)

question

In SHM, the speed of the block is zero when it is ...
a. in the equilibrium position.
b. at either its rightmost or leftmost position.
d. between its rightmost position and the equilibrium position.
d. between its leftmost position and the equilibrium position.

answer

b. (at either its rightmost or leftmost position.)

question

In SHM, the speed of the block reaches its maximum value when the block is ...
a. in the equilibrium position.
b. at either its rightmost or leftmost position.
d. between its rightmost position and the equilibrium position.
d. between its leftmost position and the equilibrium position.

answer

a. (in the equilibrium position.)

question

Because of the periodic properties of SHM, the mathematical equations that describe this motion involve sine and cosine functions. For example, if the block is released at a distance A from its equilibrium position, its displacement x varies with time t according to the
equation: x = Acoswt
where w is a constant characteristic of the system. If time is measured is seconds, w must be expressed in radians per second so that the quantity is expressed in radians.
Use this equation and the information you now have on the acceleration and speed of the block as it moves back and forth from one side of its equilibrium position to the other to determine the correct set of equations for the block's x components of velocity and acceleration, v_x and a_x, respectively. In the expressions below, B and C are nonzero positive constants.
a. v_x = βB sin Οt, a_x = C cos Οt
b. v_x = B cos Οt, a_x = C sin Οt
c. v_x = βB cos Οt, a_x = βC cos Οt
d. v_x = βB sin Οt, a_x = βC cos Οt

answer

d. (v_x = βB sin Οt, a_x = βC cos Οt)

question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that the pendulum completes 99.0 full swing cycles in a time of 139s.
What is the magnitude of the gravitational acceleration on this planet?
g_planet = ___ m/s^2

answer

T = 2 β(l/g)
T/(2 ) = β(l/g)
(T/(2 ))^2 = (l/g)
g = l / (T/(2 ))^2
T = t/#cycles => T = 139/99 = 1.40
l = 54.0 cm => 0.54 m
g = (0.54 m) / ( (1.40) / (2 ) )^2
g_planet = 10.81 m/s^2

question

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T.
If the bob's mass is doubled, approximately what will the pendulum's new period be?

answer

T = 2 β(L/g)
Since period doesn't depend on mass, it will stay the same.
T_new = T_original

question

A simple pendulum consisting of a bob of mass m attached to a string of length L swings with a period T.
If the pendulum is brought on the moon where the gravitational acceleration is about 1/6, approximately what will its period now be?

answer

T = 2 β(L/g) (since 2 and L stay the same, we can represent both as a 1 and express T as a proportionality instead)
T β β(1/g) (considering g = 1/6)
T β β(1/ (1/6) ) => T β β6
T_new = β6T

question

If a simple pendulum is taken into the orbiting space station what will happen to the bob?
a. It will continue to oscillate in a vertical plane with the same period.
b. It will no longer oscillate because there is no gravity in space.
c. It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
d. It will oscillate much faster with a period that approaches zero.

answer

c. (It will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.)

question

A 50.0-g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m. It is released with an amplitude 0.300m. A damping force F_x = -bv acts on the egg. After it oscillates for 5.00s, the amplitude of the motion has decreased to 0.100m.
Calculate the magnitude of the damping coefficient b.

answer

m = 50g = 0.05kg, k = 25.0 N/m, t_1 = 0.0 s, A_1 = 0.30m, t_2 = 5.0s, A_2 = 0.10m
x(t) = Ae^(-rt)cos(wt+Ο) (x(t)= amplitude at that time, A = initial amplitude)
@ t_f = 5.0s: x(5.0) = (0.3)e^(-r(5.0))cos(w'(5)+Ο)
( [cos(w't+Ο)] is irrelevant since we only care about amplitude [Ae^(-rt)], also the cosine function goes to 1)
=> x(5.0) = 0.1 = (0.3)e^(-5.0r) => 1/3 = e^(-5.0r) => ln(1/3) = -5.0r
***r=b/2m
ln(1/3) = -5.0(b/2m) (m=.05kg)
==> ln(1/3) = -5.0(b/2(.05)
==> b = 0.022 kg/s

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
At what time t_1 does the block come back to its original equilibrium position (x=0) for the first time?
Express your answer in terms of some or all of the variables: A, k, and m.

answer

x(t) = Acos(β(k/m) t)
0 = Acos(β(k/m) t) (divide by A)
0 = cos(β(k/m) t)
[ cos^(-1) (0) = 90Β° (convert 90Β° to radians) 90Β° * /180 = /2]
=> /2 = β(k/m) t
=> t_1 = /(2β(k/m))

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
Find the velocity of the block as a function of time.
Express your answer in terms of some or all of the variables: k, m, and A.

answer

v(t) = dx/dt = d/dt [x(t)] = d/dt [ Acos(β(k/m) t) ]
=> A (d/dt [cos(β(k/m) t)]) (d/dt [β(k/m) t]) (chain rule)
=> A( -sin(β(k/m) t) )( β(k/m) )
=> v(t) = -A(β(k/m))sin(β(k/m) t)

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
What is the maximum speed v_max of the block during this motion?
Express your answer in terms of some of all of the variables: k, m, and A.

answer

velocity at any time is expressed by the equation:
v(t) = -A(β(k/m))sin(β(k/m) t)
Since the equation contains a trig function, v_max occurs when sin(Ξ) = 1 (Ξ = /2). Therefore, sin(β(k/m) t) = 1 at v_max so we can leave it out.
v = -A(β(k/m))
Furthermore, since v_max implies that it's a positive number or a magnitude, the negative sign is irrelevant and we can leave it out of the equation.
=> v_max = A(β(k/m))

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
Find the acceleration 'a' of the block as a function of time.
Express your answer in terms of some of all of the variables: k, m, and A.

answer

a(t) = d^(2)x/dt^(2) = dv/dt = d/dt (v(t))
=> d/dt [-A(β(k/m))sin(β(k/m) t)]
=> -A(β(k/m)) (d/dt [sin(β(k/m) t)]) (d/dt [β(k/m) t]) (chain rule)
=> -A(β(k/m))(cos(β(k/m) t) )( β(k/m) )
=> a(t) = -A(k/m)(cos(β(k/m) t) )

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
Specify when the magnitude of the acceleration of the block reaches its maximum value. Consider the following options and pick which ones are true.
a. only once during one period of motion
b. when the block's speed is zero
c. when the block is in the equilibrium position
d. when the block's displacement equals either A or -A
e. when the block's speed is at a maximum

answer

b (when the block's speed is zero) and d (when the block's displacement equals either A or -A)

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
Find the kinetic energy K of the block as a function of time.
Express your answer in terms of some or all of the variables: k, m, A, and t.

answer

K = mv^2
K = 1/2 m[A(β(k/m))sin(β(k/m) t)]^2
K = 1/2 (m) (A^2) (k/m) sin^2 (β(k/m) t)
=> K(t) = 1/2 A^(2)k sin^2 (β(k/m) t)

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
Find K_max, the maximum kinetic energy of the block.
Express your answer in terms of some or all of the variables: k, m, and A.

answer

K_max when x=0, when velocity is at it's maximum.
We already found V_max to be = A(β(k/m)) and K(t) = 1/2 A^(2)k sin^2 (β(k/m) t).
(1st way) Like the trig function in V_max, we can let sin^2 (β(k/m) t) = 1 in the kinetic energy expression we found.
=> K_max = 1/2 A^(2)k
(2nd way) Since K = 1/2 mv^2, we can substitute V_max into v to get K_max.
=> K_max = 1/2 A^(2)k

question

A block of mass m is attached to a spring whose spring constant is k. The other end of the spring is fixed so that when the spring is unstretched, the mass is located at x=0. Assume that the +x direction is to the right.
The mass is now pulled to the right a distance A beyond the equilibrium position and released, at time t=0, with zero
initial velocity.
Assume that the vertical forces acting on the block balance each other and that the tension of the spring is, in effect, the
only force affecting the motion of the block. Therefore, the system will undergo simple harmonic motion. For such a
system, the equation of motion is
a(t) = -(k/m) x(t)
and its solution, which provides the equation for x(t), is
x(t) = Acos(β(k/m) t)
The kinetic energy of the block reaches its maximum when which of the following occurs?
a. The displacement of the block is zero.
b. The displacement of the block is A.
c. The acceleration of the block is at a maximum.
d. The velocity of the block is zero.

answer

a. (The displacement of the block is zero)

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and
x_init.
It is known that a general solution for the position of a harmonic oscillator is x(t) = C cos(wt) + S sin(wt)
where C, S, and w are constants.
Your task, therefore, is to determine the values of C, S, and w in terms of k, m, and x_init and then use the connection between x(t) and a(t) to find the acceleration.
Combine Newton's 2nd law and Hooke's law for a spring to find the acceleration of the block a(t) as a function of time.
Express your answer in terms of k, m, and the coordinate of the block x(t).

answer

Newtons second law is F = ma(t) and Hooks law is F = -kx(t).
solve Newtons second law for a(t):
a(t) = F/m
Substitute -kx(t) for F according to Hookes law:
=> a(t) = -(kx(t))/m

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and
x_init.
It is known that a general solution for the position of a harmonic oscillator is x(t) = C cos(wt) + S sin(wt)
where C, S, and w are constants.
Your task, therefore, is to determine the values of C, S, and w in terms of k, m, and x_init and then use the connection between x(t) and a(t) to find the acceleration.
Using the fact that acceleration is the second derivative of position, find the acceleration of the block as a function of time.
Express your answer in terms of w, t, and x(t).

answer

a(t) = dv/dt = d^2 x/dt^2 = d^2/dt^2 [C cos(wt) + S sin(wt)]
v(t) = -Cw sin(wt) + Sw cos(wt)
a(t) = -Cw^2 cos(wt) - Sw^2 sin(wt)
@ v_initial = 0 => v(0) = 0
0 = -Cw sin(w0) + Sw cos(w0) => 0 = Sw(1)
***w can't be zero because it would make x(t) = 0 for any time. Therefore, S=0
@ x_init = x(0)
x_init = C cos(w0) + S sin(w0)
x_init = C(1) => x_init = C
Plugging C and S into a(t) & x(t)
a(t) = -(x_init)w^2 cos(wt) - (0)w^2 sin(wt)
=> a(t) = -(x_init)w^2 cos(wt)
x(t) = (x_init)cos(wt) + (0)sin(wt)
=> x(t) = (x_init)cos(wt)
We can substitute x(t) for (x_init)cos(wt) in a(t)
=> a(t) = -w^2 x(t)

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the acceleration of the block a(t) as a function of time in terms of k, m, and
x_init.
It is known that a general solution for the position of a harmonic oscillator is x(t) = C cos(wt) + S sin(wt)
where C, S, and w are constants.
Your task, therefore, is to determine the values of C, S, and w in terms of k, m, and x_init and then use the connection between x(t) and a(t) to find the acceleration.
Find the angular frequency w.
Express your answer in terms of k and m.

answer

Using hooks and newtons law, we found a(t) = -k/m * x(t)
=> a(t) = -k/m (C cos(wt) + S sin(wt))
using the fact the a(t) = d^2x/dt^2 we found a(t) = -w^2 x(t) => a(t) = -w^2 (C cos(wt) + S sin(wt))
Since these expressions are the same, -k/m must equal -w^2. Use this to solve for w
-w^2 = -k/m
=> w = βk/m

question

Consider two massless springs connected in series. Spring 1 has a spring constant k_1, and spring 2 has a spring constant k_2. A constant force of magnitude F is being applied to the right. When the two springs are connected in this way, they form a system equivalent to a single spring of spring constant k.
What is the effective spring constant k of the two-spring system?
Express the effective spring constant in terms of k_1 and k_2.

answer

According to hookes law:
F = k_1x_1 = F = k_2x_2
Solve for get k_1 and k_2 on one side of equation to get effective spring constant.
x_1 + x_2 = F/k_1 + F/k_2
=> x_1 + x_2 = F(1/k_1 + 1/k_2)
=> (x_1 + x_2)/F = 1/k_1 + 1/k_2
***Since k = F/x, equation needs to be inverted
F/(x_1 + x_2) = k_1 + k_2
*** k = F/(x_1 + x_2)
=> k = k_1 + k_2

question

Consider two massless springs connected in parallel. Springs 1 and 2 have spring constants k_1 and k_2 and are connected via a thin, vertical rod. A constant force of
magnitude F is being exerted on the rod. The rod remains perpendicular to the direction of the applied force, so that the springs are extended by the same amount.
This system of two springs is equivalent to a single spring, of spring constant k.
Find the effective spring constant k of the two-spring system.
Give your answer for the effective spring constant in terms of k_1 and k_2.

answer

The system describes springs connected in parallel which act as one spring. Therefore, the force can be expressed as
F = F_1 + F_2
Applying Hooke's law we get:
F = -k_1 x - k_2 x => F= -(k_1 + k_2) x
Since we are told that the springs act as one:
F = - k_tot x
substituting - k_tot x for F...
- k_tot x = -(k_1 + k_2) x
=> k_effective = k_1 + k_2

question

Now consider three springs connected in parallel. The spring constants of springs 1, 2, and 3 are k_1, k_2, and k_3.
The springs are connected by a vertical rod, and a force of magnitude F is being exerted to the right.
Find the effective spring constant k' of the three-spring system.
Give your answer in terms of k_1, k_2, and k_3.

answer

k' = k_1 + k_2 + k_3

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the position of the block x(t) as a function of time in terms of w and
x_init.
It is known that a general solution for the position of a harmonic oscillator is
x(t) = C cos(wt) + S sin(wt) where C, S, and w are constants.
Using the general equation x(t) for given in the problem introduction, express the initial position of the block x_init in terms of C, S, and w.

answer

@ x_init = x(0)
x_init = C cos(w0) + S sin(w0)
x_init = C(1)
=> x_init = C

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the position of the block x(t) as a function of time in terms of w and
x_init.
It is known that a general solution for the position of a harmonic oscillator is
x(t) = C cos(wt) + S sin(wt) where C, S, and w are constants.
Find the value of S using the given condition that the initial velocity of the block is zero: v(0) = 0

answer

1. Find v(t) by d/dt [x(t)]
v(t) = d/dt [C cos(wt) + S sin(wt)]
==> v(t) = -Cw sin(wt) + Sw cos(wt)
2. Evaluate @ t=0
given v(0) = 0
=> 0 = -(x_init)w sin(w0) + Sw cos(w0)
=> 0 = 0 + Sw (1) => 0 = Sw
***w can't be zero because it would make x(t) = 0 for any time. Therefore, S=0

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the position of the block x(t) as a function of time in terms of w and
x_init.
It is known that a general solution for the position of a harmonic oscillator is
x(t) = C cos(wt) + S sin(wt) where C, S, and w are constants.
What is the equation x(t) for the block?
Express your answer in terms of t, w, and x_init.

answer

1. plug in values found for C and S into the given x(t)
x(t) = x_init cos(wt) + 0sin(wt)
=> x(t) = x_init cos(wt)

question

One end of a spring with spring constant k is attached to the wall. The other end is attached to a block of mass m. The block rests on a frictionless horizontal surface. The equilibrium position of the left side of the block is defined to be x=0. The length of the relaxed spring is L. The block is slowly pulled from its equilibrium position to some position
x_init > 0 along the x axis. At time t=0, the block is released with zero initial velocity.
The goal of this problem is to determine the position of the block x(t) as a function of time in terms of w and
x_init.
It is known that a general solution for the position of a harmonic oscillator is
x(t) = C cos(wt) + S sin(wt) where C, S, and w are constants.
Now, imagine that we have exactly the same physical situation but that the x axis is translated, so that the position of the wall is now defined to be x=0.
The initial position of the block is the same as before, but in the new coordinate system, the block's starting position is
given by x_new(t=0) = L + x_init
Find the equation for the block's position x_new(t) in the new coordinate system.
Express your answer in terms of L, x_init, w, and t.

answer

1. plug in x_init into the equation for x_new
x_new (t) = L + x_init cos(wt)

question

A block of mass m is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h from its
equilibrium length. The spring has an unknown spring constant k.
What is the spring constant k?
Express the spring constant in terms of given quantities and g, the magnitude of the acceleration due to gravity.

answer

1. Use hooke's and newton's law to define the oscillator
F=-kx & F=mg
-kx = -mg (negative mg b/c direction of gravity)
***|| F_spring || = || F_gravity || (force from spring and gravity must be equal and opposite)
2. solve for k
-k = - mg/x
=> k = (mg)/x

question

A block of mass m is attached to the end of an ideal spring. Due to the weight of the block, the block remains at rest when the spring is stretched a distance h from its
equilibrium length. The spring has an unknown spring constant k.
Suppose that the block gets bumped and undergoes a small vertical displacement. Find the resulting angular frequency w of the block's oscillation about its equilibrium
position.
Express the frequency in terms of given quantities and g, the magnitude of the acceleration due to gravity.

answer

1. Substitute formula in first part for k into the formula for w
w = β(k/m) ***k=mg/x
=> w = β( [mg/x]/m)
=> w = β(g/x) (note: problem uses variable h for x)

question

The steel used for piano wire has a tensile strength p_T of 3*10^9 N/m^2 and a density β΄ of 7800 kg/m^3.
What is the speed v of a wave traveling down such a wire if the wire is stretched to its breaking point?
express the speed in m/s to the nearest integer.

answer

p_T = 3*10^9 N/m^2, β΄ = 7800 kg/m^3
1. Substitute F_T=(p_T)A and ΞΌ=β΄A for F_T and ΞΌ in the equation v = β(F_T/ΞΌ)
=> v = β( (p_T)A)/(β΄A) )
=> v = β(p_T/β΄)
=> v = β(3*10^9/7800)
=> v = 620 m/s

question

The steel used for piano wire has a tensile strength p_T of 3*10^9 N/m^2 and a density β΄ of 7800 kg/m^3.
If this wire is used for high C (4000 Hz), what must the vibrating length be?
Express the length in centimeters to three significant figures.

answer

1. note that for the piano to play high C in tune, it must be in the fundamental frequency. Therefore, we can use the equation f = n (v/2L) where n=1
=> f = v/2L
=> L = v/2f
=> L = 620/2(4000)
=> L = .0775 m => 7.75 cm

question

A standing wave can be mathematically expressed as
y(x,t) = Asin(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
Which one of the statements is correct?
a. wave is traveling in +x direction
b. wave is traveling in -x direction
c. wave is oscillating but not traveling
d. wave is traveling but not oscillating

answer

c. (wave is oscillating but not traveling)

question

A standing wave can be mathematically expressed as
y(x,t) = Asin(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
At time t=0, what is the displacement of the string y(x,0)?
Express your answer in terms of A, k, and other introduced quantities.

answer

1. just plug in t=0 into the given equation
@t=0: y(x,0) = Asin(kx)sin(w(0))
=> y(x,0) = Asin(kx)(0)
=> y(x,0) = 0

question

A standing wave can be mathematically expressed as
y(x,t) = Asin(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
What is the displacement of the string as a function of x at time T/4 where T is the period of oscillation?
Express your answer in terms of A, k, and other introduced quantities.

answer

1. Evaluate w(T/4)
w = 2 /T
=> w = 2 /(T/4)
=> w = /(2T)
=> wt = /2
2. substitute /2 into y(x,t).
y(x, T/4) = Asin(kx)sin( /2)
=> y(x, T/4) = Asin(kx)

question

A standing wave can be mathematically expressed as
y(x,t) = Asin(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
At which points x1, x2, and x3 closest to x=0 but with x>0 will the displacement of the string y(x,t) be zero for all times?
Express x1, x2, and x3 in terms of wavelength.

answer

1. Note that at the nodes, y(x,t) = 0. So we can set the expression for y(x,t) equal to 0.
0 = Asin(kx)sin(wt)
2. Since Asin(kx) expresses wavelength, everything else is irrelevant and can be ignored.
0 = Asin(kx)
kx = sin^-1 (0)
3. sin(Ξ) = 0 at , 2 , and 3 so we can set kx equal to these values and solve for x. But we need to express k as K=2 /Ξ»
kx = Ξ => (2 /Ξ») x = Ξ => x = (1/2 )ΞΞ»
: x1 = (1/2 )( )Ξ» => x1 = 1/2 Ξ»
2 : x2 = (1/2 )(2 )Ξ» => x2 = 1 Ξ»
3 : x3 = (1/2 )(3 )Ξ» => x3 = 2/3 Ξ»

question

A standing wave can be mathematically expressed as
y(x,t) = Acos(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
Which one of the statements is correct?
a. wave is traveling in +x direction
b. wave is traveling in -x direction
c. wave is oscillating but not traveling
d. wave is traveling but not oscillating

answer

c. (wave is oscillating but not traveling)

question

A standing wave can be mathematically expressed as
y(x,t) = Acos(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
At time t=0, what is the displacement of the string y(x,0)?
Express your answer in terms of A, k, and other introduced quantities.

answer

1. just plug in t=0 into the given equation
@t=0: y(x,0) = cos(kx)sin(w(0))
=> y(x,0) = Asin(kx)(0)
=> y(x,0) = 0

question

A standing wave can be mathematically expressed as
y(x,t) = Acos(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
What is the displacement of the string as a function of x at time T/4 where T is the period of oscillation?
Express your answer in terms of A, k, and other introduced quantities.

answer

1. Evaluate w(T/4)
w = 2 /T
=> w = 2 /(T/4)
=> w = /(2T)
=> wt = /2
2. substitute /2 into y(x,t).
y(x, T/4) = Acos(kx)sin( /2)
=> y(x, T/4) = Acos(kx)

question

A standing wave can be mathematically expressed as
y(x,t) = Acos(kx)sin(wt)
A = max transverse displacement (amplitude), k = wave number, w = angular frequency, t = time.
At which points x1, x2, and x3 closest to x=0 but with x>0 will the displacement of the string y(x,t) be zero for all times?
Express x1, x2, and x3 in terms of wavelength.

answer

1. Note that at the nodes, y(x,t) = 0. So we can set the expression for y(x,t) equal to 0.
0 = Acos(kx)sin(wt)
2. Since Acos(kx) expresses wavelength, everything else is irrelevant and can be ignored.
0 = Acos(kx)
kx = cos^-1 (0)
3. cos(Ξ) = 0 at /2, 3 /2, and 5 /2 so we can set kx equal to these values and solve for x. But we need to express k as K=2 /Ξ»
kx = Ξ => (2 /Ξ») x = Ξ => x = (1/2 )ΞΞ»
/2: x1 = (1/2 )( /2)Ξ» => x1 = 1/4 Ξ»
3 /2: x2 = (1/2 )(3 /2)Ξ» => x2 = 3/4 Ξ»
5 /2: x3 = (1/2 )(5 /2)Ξ» => x3 = 5/4 Ξ»

question

Four waves are described by the expressions:
a. y = 0.12 cos(3x - 21t)
b. y = 0.15 sin(6x + 42t)
c. y = 0.13 cos(6x + 21t)
d. y = -0.23 sin(3x - 42t)
Which two waves have the same speed?

answer

Notice each equation is in the form: Acos(kx -wt). Therefore, we may use the k and w values from each expression and use the formula w = kv (from the help sheet) to calculate velocity.
w = vk => v = w/k
a. w = 21, k = 3: v = 21/3 => 7
b. w = 42, k = 6: v = 42/6 => 7
*** therefore, we already know a and b have the same speed and we dont necessarily have to solve the rest (prolly should though)