Question 1

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = -1/2x_m, the kinetic and potential energies are: A) K = 5J and U = 3J B) K = 5J and U = -3J C) K = 8J and U = 0 D) K = 0 and U = 8J E) K = 0 and U = -8J

Accepted Answer

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

Question 2

A certain spring elongates 9 mm when it is suspended vertically and a block of mass M is hung on it. The natural frequency of this mass-spring system is:
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 natural frequency of a mass-spring system can be calculated as:
 
ω0 = √(k/M)

where k is the spring constant.

To find k, we can use Hooke's Law:

F = -kx

where F is the force applied on the spring, x is the displacement, and the negative sign indicates that the force is in the opposite direction of the displacement.

In this case, the elongation of the spring is x = 9 mm = 0.009 m, and the weight of the block is F = Mg, where g is the acceleration due to gravity. Therefore:

k = F/x = Mg/x

Substituting this into the equation for ω0, we get:

ω0 = √(Mg/xM) = √(g/x)

Using the value of x, we get:

ω0 = √(9.81/0.009) = 33 rad/s

Therefore, the best choice is B. Note that the mass of the block (M) is not required to calculate the natural frequency.

Question 3

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) zero
B) 12.5 J
C) 25 J
D) 37.5 J
E) 50 J

Accepted Answer

The total energy of the spring block system is the sum of its potential energy and kinetic energy:
E = 1/2 kA^2 + 1/2 mv^2
where k is the spring constant, A is the amplitude, and v is the velocity of the block.
At half the amplitude, the displacement x = A/2. The potential energy at this point is:
U = 1/2 k (A/2)^2 = 1/8 kA^2
Since the total energy is 50 J, the kinetic energy at this point is:
K = E - U = 50 - 1/8 kA^2
We don't know the values of k and A, but we can still use the information given in the question to find the answer. Since the motion is simple harmonic, the velocity and displacement are related by:
v^2 = k/m (A^2 - x^2)
At x = A/2,
v^2 = k/m (A^2 - A^2/4) = 3/4 kA^2/m
The kinetic energy at this point is therefore:
K = 1/2 mv^2 = 1/2 (3/4 kA^2/m) = 3/8 kA^2
Substituting this into the expression for K above, we get:
3/8 kA^2 = 50 - 1/8 kA^2
Solving for kA^2, we get:
kA^2 = 200/5 = 40
Therefore, the kinetic energy at half the amplitude is:
K = 3/8 kA^2 = 3/8 * 40 = 15 J
Since the answer choices are in increments of 12.5 J, the closest option is D: 37.5 J.

Question 4

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

The equation of motion for a simple harmonic oscillator is given by $x = A\cos(\omega t)$, where $A$ is the amplitude of motion, $\omega$ is the angular frequency, and $t$ is time. In this case, $A = 2$ m and $\omega = 50$ rad/s. The angular frequency $\omega$ is related to the mass $m$ and the spring constant $k$ by the formula $\omega = \sqrt{\frac{k}{m}}$. Rearranging this formula to solve for $k$ gives $k = m\omega^2$. Substituting $m = 3$ kg and $\omega = 50$ rad/s into this formula yields $k = 3 \times (50)^2 = 3 \times 2500 = 7500$ N/m.

Question 5

It is impossible for two particles, each executing simple harmonic motion, to remain in phase with each other if they have different:
A) masses
B) periods
C) amplitudes
D) spring constants
E) kinetic energies

Accepted Answer

The period of a simple harmonic motion is dependent only on the mass and spring constant of the oscillator. Therefore, if two particles have different periods, it means they have different mass or spring constant, making it impossible for them to remain in phase with each other. The other factors (amplitude and kinetic energy) do not affect the period or phase relationship between the particles.

Question 6

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

In simple harmonic motion, the restoring force is always proportional to the displacement from equilibrium (Hooke's law). This means that as the displacement increases, the restoring force also increases proportionally. The other options are not true because amplitude and frequency do not directly affect the restoring force, and velocity and displacement squared are not directly proportional to the restoring force.

Question 7

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

For a simple harmonic oscillator, the total mechanical energy (the sum of kinetic and potential energy) is constant over a cycle. Therefore, K_avg = U_avg = constant. Options A and B imply that one average energy is greater than the other, which is not possible since they are equal. Option D is not always true since the oscillator can have non-zero kinetic energy when its displacement is zero. Option E is also not true in general for a simple harmonic oscillator since the energies can take any positive value (as long as their sum is constant).

Question 8

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 maximum displacement, or amplitude, of the oscillation can be found using the formula:

A = (2E/mω^2)^0.5

where E is the initial mechanical energy of the system, m is the mass of the block, and ω is the angular frequency of the oscillation.

To find E, we can use the conservation of energy:

0.5kx^2 + 0.5mv^2 = 0.5kA^2

where x is the initial elongation of the spring, v is the initial speed of the block, and A is the amplitude.

Plugging in the given values, we get:

0.5 (200 N/m) (0.15 m)^2 + 0.5 (0.25 kg) (3.0 m/s)^2 = 0.5 (200 N/m) A^2

Solving for A, we get:

A = 0.18 m

Therefore, the answer is B.

Question 9

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 B) $\pi$/2 radians C) $\pi$ radians D) 3 $\pi$/2 radians E) 2 $\pi$ radians

Accepted Answer

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Question 10

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

The maximum speed occurs at the equilibrium position (where the spring is neither compressed nor extended), and is equal to the amplitude times the angular frequency:

v_max = A*omega

The angular frequency can be found from the mass and spring constant:

omega = sqrt(k/m)

Putting it all together, solving for the amplitude:

A = v_max/omega = v_max*sqrt(m/k) = 5.0 m/s * sqrt(0.20 kg / 500 N/m) = 0.10 m

Question 11

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 answer of An object on the end of a...

Question 12

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 13

A simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. The particle oscillates as shown in (i) with period T. If the spring is cut in half and used with the same particle, as shown in (ii), the period will be: 
A) 2T
B) 
C) 
D) T
E) T/2

Accepted Answer

The answer of A simple harmonic oscillator consists of a...

Question 14

An object of mass m, oscillating on the end of a spring with spring constant k has amplitude A. Its maximum speed is: A) B) A²k/m C) D) Am/k E) A²m/k

Accepted Answer

The answer of An object of mass m, oscillating on...

Question 15

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, it means that the object is at its maximum displacement at t = 0 and moving in the negative x direction at t = 0. From the given equation, we see that the cosine term is at its maximum value when the argument is 0 or a multiple of 2π. Therefore, we have: x(0) = x_mcos(F1F1F1S1F1F1F10) = -x_m x'(0) = -x_mF1F1F1S1F1F1F10sin(F1F1F1S1F1F1F10) < 0 From these conditions, we know that cos(F1F1F1S1F1F1F10) = -1 and sin(F1F1F1S1F1F1F10) < 0. This occurs in the second quadrant of the unit circle where the angle is between π/2 and π. Therefore, we have: π/2 < F1F1F1S1F1F1F10 < π F1F1F1S1F1F1F10/2 < F1F1F1S1F1F1F10 < F1F1F1S1F1F1F10 Therefore, the correct choice is B, F1F1F1S1F1F1F10/2 to F1F1F1S1F1F1F10 radians.

Question 16

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 17

A particle moves in simple harmonic motion according to x = 2cos(50t), where x is in meters and t is in seconds. Its maximum velocity in m/s is:
A) 100 sin(50t)
B) 100 cos(50t)
C) 100
D) 200
E) none of these

Accepted Answer

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

Question 18

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 19

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

Accepted Answer

In simple harmonic motion, acceleration and velocity are out of phase by π/2 or F1F1F1S1F1F1F10/2 rad. Therefore, the best choice is option D.

Question 20

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = x_m, the kinetic and potential energies are: A) K = 5J and U = 3J B) K = 5J and U = -3J C) K = 8J and U = 0 D) K = 0 and U = 8J E) K = 0 and U = -8J

Accepted Answer

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

Quiz 15: Oscillations

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = -1/2x_m, the kinetic and potential energies are:

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

It is impossible for two particles, each executing simple harmonic motion, to remain in phase with each other if they have different:

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

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 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:

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 0.20-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:

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 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 simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. The particle oscillates as shown in (i) with period T. If the spring is cut in half and used with the same particle, as shown in (ii) , the period will be:

An object of mass m, oscillating on the end of a spring with spring constant k has amplitude A. Its maximum speed 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:

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 particle moves in simple harmonic motion according to x = 2cos(50t) , where x is in meters and t is in seconds. Its maximum velocity in m/s is:

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:

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

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = x_m, the kinetic and potential energies are:

Quiz 15: Oscillations

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = -1/2xm, the kinetic and potential energies are:

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

It is impossible for two particles, each executing simple harmonic motion, to remain in phase with each other if they have different:

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

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:

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:

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 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:

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 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 simple harmonic oscillator consists of a mass m and an ideal spring with spring constant k. The particle oscillates as shown in (i) with period T. If the spring is cut in half and used with the same particle, as shown in (ii) , the period will be:

An object of mass m, oscillating on the end of a spring with spring constant k has amplitude A. Its maximum speed is:

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:

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 particle moves in simple harmonic motion according to x = 2cos(50t) , where x is in meters and t is in seconds. Its maximum velocity in m/s is:

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:

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

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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = xm, 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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = -1/2x_m, the kinetic and potential energies are:

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:

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:

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 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 = 5J and its potential energy (measured with U = 0 at x = 0) is U = 3J. When it is at x = x_m, the kinetic and potential energies are: