Question 1

A 0.20-kg object mass attached to a spring whose spring constant is 500 N/m executes simple harmonic motion. If its maximum speed is 5.0 m/s, the amplitude of its oscillation is:
A) 0.0020 m
B) 0.10 m
C) 0.20 m
D) 25 m
E) 250 m

Accepted Answer

First, we need to find the maximum potential energy of the system. This occurs when the spring is at its maximum displacement (which we will call "A", the amplitude). The maximum potential energy can be found using the formula:

U = (1/2)kA^2

where k is the spring constant.

Next, we need to find the maximum kinetic energy of the system. This occurs when the object is at the equilibrium position (i.e. the midpoint between its maximum displacement in either direction). At this point, all of the potential energy has been converted to kinetic energy. The maximum kinetic energy can be found using the formula:

K = (1/2)mv^2

where m is the mass of the object and v is the maximum speed.

Finally, we can use the fact that the total mechanical energy of the system (i.e. the sum of the potential and kinetic energies) is constant to solve for the amplitude:

U + K = (1/2)kA^2 + (1/2)mv^2 = constant

So, rearranging this equation, we get:

A = sqrt[(2/m)(constant - (1/2)mv^2)]

Plugging in the values given in the problem, we get:

A = sqrt[(2/0.20)(0.5*k*A^2 + 0.5*m*v^2)]

Simplifying this expression, we get:

A = sqrt[(5*k/m)v^2]

Plugging in the values given in the problem, we get:

A = sqrt[(5*500/0.20)*25] = 0.10 m

Therefore, the answer is B.

Question 2

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the amplitude of the oscillation is:
A) 0.13 m
B) 0.18 m
C) 3.7 m
D) 5.2 m
E) 13 m

Accepted Answer

The total energy of the system at the beginning of the oscillation is the sum of the kinetic and potential energy:
E = 1/2 mv^2 + 1/2 kx^2
where m is the mass of the block, v is its initial velocity, k is the spring constant and x is the initial elongation of the spring. Plugging in the given values, we get:
E = 1/2 (0.25 kg) (3.0 m/s)^2 + 1/2 (200 N/m) (0.15 m)^2 = 1.35 J
At the ends of the oscillation, all the energy is in the form of potential energy in the spring. The amplitude A is the maximum elongation of the spring, when all the energy momentarily transforms between kinetic and potential forms. Thus, at the ends of the oscillation, we have:
1/2 kA^2 = E
Solving for A, we get:
A = sqrt(2E/k) = sqrt(2(1.35 J)/(200 N/m)) = 0.18 m

Question 3

Let U be the potential energy (with the zero at zero displacement) and K be the kinetic energy of a simple harmonic oscillator. U_avg and K_avg are the average values over a cycle. Then: A) K_avg > U_avg B) K_avg < U_avg C) K_avg = U_avg D) K = 0 when U = 0 E) K + U = 0

Accepted Answer

In a simple harmonic motion, the potential energy and kinetic energy are both oscillatory and at any point in time, the sum of the two energies remains constant. Therefore, if K_avg represents the average kinetic energy over a cycle and U_avg represents the average potential energy over a cycle, then K_avg = U_avg, as both will have the same average value over the cycle. Option D is incorrect, as the kinetic energy will never be zero unless the oscillator is at rest. Option E is wrong because the sum of kinetic energy and potential energy remains constant, but not necessarily zero.

Question 4

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $ \omega $t + $\phi$ ). If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant $\phi$ is: A) 0 radians B) $\pi$/2 radians C) $\pi$ radians D) 3 $\pi$/2 radians E) 2 $\pi$ radians

Accepted Answer

Since the initial velocity is in the negative x direction, we know that the object starts by moving towards the equilibrium point with a negative velocity. This means that at time t=0, the object is located at the maximum displacement in the negative direction. Therefore, we can write:

x(0) = -xS1

Plugging this into our equation for x(t), we get:

-xS1 = xS1 U1B1mS1U1B0 cos(F1F1F1S1F1F1F10)

We can simplify this to:

cos(F1F1F1S1F1F1F10) = -1

Which means that the phase constant F1F1F1S1F1F1F10 must be an odd multiple of pi/2. The only answer choice that satisfies this condition is B, F1F1F1S1F1F1F10/2 radians.

Question 5

A certain spring elongates 9 mm when it is suspended vertically and a block of mass M is hung on it. The angular frequency of this mass-spring system:
A) is 0.088 rad/s
B) is 33 rad/s
C) is 200 rad/s
D) is 1140 rad/s
E) cannot be computed unless the value of M is given

Accepted Answer

The spring-mass system can be modeled using Hooke's Law and the equation for simple harmonic motion.

First, we can use Hooke's Law to find the spring constant:
F = kx
where F is the force exerted by the spring, x is its displacement, and k is the spring constant.

In this case, the displacement is 9 mm and the force is the weight of the block:
F = mg
where g is the acceleration due to gravity (9.81 m/s^2).

Thus, we have:
kx = mg
k = mg/x
where m is the mass of the block.

Substituting in the values given:
k = (M)(9.81 m/s^2)/(9 mm) = 1078.33 N/m

Next, we can use the equation for simple harmonic motion:
x = A sin(ωt)
where x is the displacement, A is the amplitude, ω is the angular frequency, and t is time.

We can use the fact that the elongation of the spring is the amplitude, A:
A = 9 mm = 0.009 m

Then, we can solve for ω:
ω^2 = k/m
ω = √(k/m)

Substituting in the values calculated:
ω = √(1078.33 N/m / M)

We can see that the angular frequency does not depend on the value of M, so we can eliminate choice E.

Calculating the value of ω:
ω = √(1078.33 N/m / M) = √(1078.33 N/m / 1 kg) = 32.879 rad/s

This matches choice B.

Question 6

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $ \omega $t + $\phi$ ). If the object is initially displaced in the negative x direction and given a negative initial velocity, then the phase constant $\phi$ is between: A) 0 and $\pi$/2 radians B) $\pi$/2 and $\pi$radians C) $\pi$ and 3 $\pi$/2 radians D) 3 $\pi$/2 and 2 $\pi$radians E) none of the above ($\phi$ is exactly 0, $\pi$/2, $\pi$, or 3 $\pi$/2 radians)

Accepted Answer

Since the object is initially displaced in the negative x direction and given a negative initial velocity, we know that the amplitude must be positive (since the distance from equilibrium is positive) and the initial phase must be between F1F1F1S1F1F1F10 and F1F1F1S1F1F1F10/2 (since it starts in the negative x direction and the cosine function is negative at F1F1F1S1F1F1F10/2). Therefore, the phase constant must be between F1F1F1S1F1F1F10/2 and F1F1F1S1F1F1F10 radians.

Question 7

In simple harmonic motion, the restoring force must be proportional to the:
A) amplitude
B) frequency
C) velocity
D) displacement
E) displacement squared

Accepted Answer

The restoring force in simple harmonic motion is proportional to the displacement from equilibrium, as given by Hooke's Law. The equation for this relationship is F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement from equilibrium. This means that as the displacement increases, the restoring force also increases. Therefore, the correct answer is D) displacement.

Question 8

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial the initial velocity is v₀ and in the second it is 4v₀. In the second trial: A) the amplitude is half as great and the maximum acceleration is twice as great B) the amplitude is twice as great and the maximum acceleration is half as great C) both the amplitude and the maximum acceleration are twice as great D) both the amplitude and the maximum acceleration are four times as great E) the amplitude is four times as great and the maximum acceleration is twice as great

Accepted Answer

The maximum velocity of an object on a spring is given by:

vmax = Aω

where A is the amplitude and ω is the angular frequency. The angular frequency is given by:

ω = sqrt(k/m)

where k is the spring constant and m is the mass of the object.

Since the mass and spring constant are constant, the angular frequency is also constant for both trials. This means that the maximum velocity in the second trial is four times greater than in the first trial:

vmax,2 = 4vmax,1

The maximum acceleration of an object on a spring is given by:

amax = Aω^2

Since ω is constant, the maximum acceleration is proportional to A^2. This means that the maximum acceleration in the second trial is 16 times greater than in the first trial:

amax,2 = 16amax,1

The amplitude in the second trial can be found by using the equation for maximum velocity:

vmax,2 = A2ω

A2 = vmax,2/ω = 4vmax,1/ω = 4A1

Therefore, the amplitude in the second trial is four times greater than in the first trial.

So, both the amplitude and the maximum acceleration are four times as great in the second trial as compared to the first trial. Therefore, the correct answer is D.

Question 9

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the maximum speed of the block is:
A) 0.06 m/s
B) 0.17 m/s
C) 0.24 m/s
D) 4.9 m/s
E) 6.9 m/s

Accepted Answer

The answer of A 0.25-kg block oscillates on the end...

Question 10

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. At one point in its motion its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = x_m, the kinetic and potential energies are: A) K = 5 J and U = 3 J B) K = 5 J and U = -3 J C) K = 8 J and U = 0 J D) K = 0 J and U = 8 J E) K = 0 J and U = -8 J

Accepted Answer

The answer of A particle is in simple harmonic motion...

Question 11

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the amplitude of the oscillation is:
A) 0.06 m
B) 0.17 m
C) 0.24 m
D) 4.9 m
E) 6.9 m

Accepted Answer

The answer of A 0.25-kg block oscillates on the end...

Question 12

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is:
A) 0 J
B) 12.5 J
C) 25 J
D) 37.5 J
E) 50 J

Accepted Answer

In simple harmonic motion, the total energy (E) is the sum of potential energy (U) and kinetic energy (K). At any point, E = U + K. The potential energy in a spring system is given by $U = \frac{1}{2}kx^2$, where $x$ is the displacement from the equilibrium position. When the displacement is half the amplitude ($x = \frac{1}{2}A$), the potential energy is $U = \frac{1}{2}k\left(\frac{1}{2}A\right)^2 = \frac{1}{8}kA^2$. Since the total energy at maximum displacement (amplitude $A$) is all potential and equals $E = \frac{1}{2}kA^2 = 50 J$, when the displacement is half, the potential energy is a quarter of the total energy ($\frac{1}{4}E = 12.5 J$), leaving $50 J - 12.5 J = 37.5 J$ for the kinetic energy.

Question 13

A 3-kg block, attached to a spring, executes simple harmonic motion according to x = 2cos(50t) where x is in meters and t is in seconds. The spring constant of the spring is:
A) 1 N/m
B) 100 N/m
C) 150 N/m
D) 7500 N/m
E) none of these

Accepted Answer

In simple harmonic motion, the equation for displacement as a function of time is given by:

x = A cos(ωt + φ)

where A is the amplitude, ω is the angular frequency, and φ is the phase angle.

Comparing this with the given equation x = 2cos(50t), we can see that:

A = 2
ω = 50 rad/s
φ = 0 (since the block is at the equilibrium position at t=0)

The angular frequency is related to the spring constant k and the mass m by the equation:

ω = sqrt(k/m)

Rearranging this equation, we get:

k = mω^2

Substituting the values of m and ω, we get:

k = 3*(50)^2 = 7500 N/m

Therefore, the spring constant of the spring is 7500 N/m, which corresponds to choice D.

Question 14

A 1.2-kg mass is oscillating without friction on a spring whose spring constant is 3400 N/m. When the mass's displacement is 7.2 cm, what is its acceleration? A) −3.8 m/s² B) −200 m/s² C) −240 m/s² D) −2.0 x 10⁴ m/s² E) cannot be calculated without more information

Accepted Answer

Difficultly: M

Question 15

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is:
A) 0.13 m/s
B) 0.18 m/s
C) 3.7 m/s
D) 5.2 m/s
E) 13 m/s

Accepted Answer

The answer of A 0.25-kg block oscillates on the end...

Question 16

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. When it is at x = x₁, its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = -1/2 x_m, the kinetic and potential energies are: A) K = 6 J and U = 2 J B) K = 6 J and U = -2 J C) K = 8 J and U = 0 J D) K = 0 J and U = 8 J E) K = 0 J and U = -8 J

Accepted Answer

At x = $x_{m}/2$, the total mechanical energy (E) is conserved and equals the sum of kinetic (K) and potential (U) energies. Initially, E = K + U = 5 J + 3 J = 8 J. At $x = -x_{m}/2$, the potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}k\left(\frac{x_m}{2}\right)^2 = \frac{1}{4}kx_m^2$. Given that at $x = x_m$, U = 8 J (total energy) - 5 J (kinetic energy at $x_{1/2}x_m$) = 3 J, and U is proportional to $x^2$, at $x = -x_{m}/2$, U = $\frac{1}{4}$ of its value at $x = x_m$, which is $\frac{1}{4} \times 8 J = 2 J$. Thus, K = E - U = 8 J - 2 J = 6 J.

Question 17

The acceleration of a body executing simple harmonic motion leads the velocity by what phase?
A) 0 rad
B)  $\pi$/8 rad
C)  $\pi$/4 rad
D)  $\pi$/2 rad
E)  $\pi$ rad

Accepted Answer

In simple harmonic motion, acceleration leads velocity by a phase angle of F1F1F1S1F1F1F10/2 rad (or 90 degrees). Therefore, the correct choice is D.

Question 18

In simple harmonic motion, the magnitude of the acceleration is greatest when:
A) the displacement is zero
B) the displacement is maximum
C) the speed is maximum
D) the force is zero
E) the speed is between zero and its maximum

Accepted Answer

The acceleration in simple harmonic motion is given by a = -ω^2x, where ω is the angular frequency and x is the displacement from equilibrium. The magnitude of acceleration is given by |a| = ω^2|x|. Since ω is a constant and |x| is greatest at maximum displacement, the magnitude of acceleration is greatest at maximum displacement.

Question 19

In simple harmonic motion:
A) the acceleration is greatest at the maximum displacement
B) the velocity is greatest at the maximum displacement
C) the period depends on the amplitude
D) the acceleration is constant
E) the acceleration is greatest at zero displacement

Accepted Answer

The answer of In simple harmonic motion:
A) the acceleration is...

Question 20

In simple harmonic motion, the magnitude of the acceleration is:
A) constant
B) proportional to the displacement
C) inversely proportional to the displacement
D) greatest when the velocity is greatest
E) never greater than g

Accepted Answer

In simple harmonic motion, the acceleration is directly proportional to the displacement (i.e., the magnitude of the displacement from the equilibrium point). This relationship is given by the equation a = -ω^2x, where a is the acceleration, x is the displacement, and ω is the angular frequency of the motion. Since ω is constant for a given system, the magnitude of the acceleration is proportional to the magnitude of the displacement.

Quiz 15: Oscillations

A 0.20-kg object mass attached to a spring whose spring constant is 500 N/m executes simple harmonic motion. If its maximum speed is 5.0 m/s, the amplitude of its oscillation is:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the amplitude of the oscillation is:

Let U be the potential energy (with the zero at zero displacement) and K be the kinetic energy of a simple harmonic oscillator. U_avg and K_avg are the average values over a cycle. Then:

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $\omega$ t + $\phi$ ) . If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant $\phi$ is:

A certain spring elongates 9 mm when it is suspended vertically and a block of mass M is hung on it. The angular frequency of this mass-spring system:

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $\omega$ t + $\phi$ ) . If the object is initially displaced in the negative x direction and given a negative initial velocity, then the phase constant $\phi$ is between:

In simple harmonic motion, the restoring force must be proportional to the:

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial the initial velocity is v₀ and in the second it is 4v₀. In the second trial:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the maximum speed of the block is:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. At one point in its motion its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = x_m, the kinetic and potential energies are:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the amplitude of the oscillation is:

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is:

A 3-kg block, attached to a spring, executes simple harmonic motion according to x = 2cos(50t) where x is in meters and t is in seconds. The spring constant of the spring is:

A 1.2-kg mass is oscillating without friction on a spring whose spring constant is 3400 N/m. When the mass's displacement is 7.2 cm, what is its acceleration?

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. When it is at x = x₁, its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = -1/2 x_m, the kinetic and potential energies are:

The acceleration of a body executing simple harmonic motion leads the velocity by what phase?

In simple harmonic motion, the magnitude of the acceleration is greatest when:

In simple harmonic motion:

In simple harmonic motion, the magnitude of the acceleration is:

Quiz 15: Oscillations

A 0.20-kg object mass attached to a spring whose spring constant is 500 N/m executes simple harmonic motion. If its maximum speed is 5.0 m/s, the amplitude of its oscillation is:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the amplitude of the oscillation is:

Let U be the potential energy (with the zero at zero displacement) and K be the kinetic energy of a simple harmonic oscillator. Uavg and Kavg are the average values over a cycle. Then:

The displacement of an object oscillating on a spring is given by x(t) = xmcos( ω \omega ω t + ϕ\phiϕ ) . If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant ϕ\phiϕ is:

A certain spring elongates 9 mm when it is suspended vertically and a block of mass M is hung on it. The angular frequency of this mass-spring system:

The displacement of an object oscillating on a spring is given by x(t) = xmcos( ω \omega ω t + ϕ\phiϕ ) . If the object is initially displaced in the negative x direction and given a negative initial velocity, then the phase constant ϕ\phiϕ is between:

In simple harmonic motion, the restoring force must be proportional to the:

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial the initial velocity is v0 and in the second it is 4v0. In the second trial:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the maximum speed of the block is:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm. At one point in its motion its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = xm, the kinetic and potential energies are:

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the system has an energy of 6.0 J, then the amplitude of the oscillation is:

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. When the displacement is half the amplitude, the kinetic energy is:

A 3-kg block, attached to a spring, executes simple harmonic motion according to x = 2cos(50t) where x is in meters and t is in seconds. The spring constant of the spring is:

A 1.2-kg mass is oscillating without friction on a spring whose spring constant is 3400 N/m. When the mass's displacement is 7.2 cm, what is its acceleration?

A 0.25-kg block oscillates on the end of the spring with a spring constant of 200 N/m. If the oscillation is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is xm. When it is at x = x1, its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = -1/2 xm, the kinetic and potential energies are:

The acceleration of a body executing simple harmonic motion leads the velocity by what phase?

In simple harmonic motion, the magnitude of the acceleration is greatest when:

In simple harmonic motion:

In simple harmonic motion, the magnitude of the acceleration is:

Let U be the potential energy (with the zero at zero displacement) and K be the kinetic energy of a simple harmonic oscillator. U_avg and K_avg are the average values over a cycle. Then:

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $\omega$ t + $\phi$ ) . If the initial displacement is zero and the initial velocity is in the negative x direction, then the phase constant $\phi$ is:

The displacement of an object oscillating on a spring is given by x(t) = x_mcos( $\omega$ t + $\phi$ ) . If the object is initially displaced in the negative x direction and given a negative initial velocity, then the phase constant $\phi$ is between:

An object on the end of a spring is set into oscillation by giving it an initial velocity while it is at its equilibrium position. In the first trial the initial velocity is v₀ and in the second it is 4v₀. In the second trial:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. At one point in its motion its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = x_m, the kinetic and potential energies are:

A particle is in simple harmonic motion along the x axis. The amplitude of the motion is x_m. When it is at x = x₁, its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0) is U = 3 J. When it is at x = -1/2 x_m, the kinetic and potential energies are: