Question 1

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 2

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 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)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 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

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 magnitude of the acceleration is given by a = -ω^2x, where ω is the angular frequency and x is the displacement. Since acceleration is proportional to displacement, it is greatest at the maximum displacement where the restoring force is strongest.

Question 6

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 7

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 it starts moving in the negative x direction. This corresponds to a phase shift of F1F1F1S1F1F1F10 radians. Also, the cosine function reaches its maximum at a phase shift of 0 radians and minimum at a phase shift of F1F1F1S1F1F1F10 radians. Therefore, the phase constant must be between F1F1F1S1F1F1F10/2 and F1F1F1S1F1F1F10 radians in order to satisfy the initial conditions.

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 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 9

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 potential energy is maximum at the extreme displacements and zero at the equilibrium position. The kinetic energy is maximum at the equilibrium position and zero at the extreme displacements. Therefore, over a cycle, the average potential energy and average kinetic energy are equal, i.e. K_avg = U_avg. Option A and B are incorrect because they suggest that one is greater than the other. Option D and E are also incorrect as they do not apply to all simple harmonic oscillators.

Question 10

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 from the equilibrium position and is directed towards that position. This relationship is described by the formula $a = -\omega^2 x$, where $a$ is the acceleration, $\omega$ is the angular frequency, and $x$ is the displacement.

Question 11

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 amplitude of the oscillation can be found using the equation for simple harmonic motion:

A = x0, where x0 is the initial displacement from equilibrium.

In this case, the initial displacement from equilibrium is the elongation of the spring, which is 0.15 m. Therefore, the amplitude of the oscillation is 0.15 m.

Note that the speed of the block is not relevant for determining the amplitude, only the initial displacement and the spring constant are needed.

Question 12

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 the equilibrium position. This is known as Hooke's Law and can be expressed as F = -kx, where F is the restoring force, k is the spring constant (a measure of how stiff the spring is), and x is the displacement from equilibrium. Therefore, the best choice is D, displacement.

Question 13

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, the acceleration leads the velocity by a phase angle of π/2 or 90 degrees. This is equivalent to a phase of "F1F1F1S1πF1F1F10/2 rad" in radians or "F1F1F1S1πF1F1F15°" in degrees. Therefore, the best choice is D, which corresponds to this value.

Question 14

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 of an object undergoing simple harmonic motion is at its equilibrium point. At this point, all of the potential energy stored in the spring is converted to kinetic energy, and the speed of the object is at its maximum. Using the equation for the maximum speed of an object in SHM, we can solve for the amplitude:

v_max = Aω

where v_max is the maximum speed, A is the amplitude, and ω is the angular frequency of the oscillation. The angular frequency can be found using the spring constant and mass of the object:

ω = √(k/m) = √(500 N/m ÷ 0.20 kg) = 22.4 rad/s

Plugging in the values given, we get:

5.0 m/s = A(22.4 rad/s)

A = 0.10 m

Therefore, the amplitude of the oscillation is 0.10 m, or choice B.

Question 15

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 16

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 greatest...

Question 17

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 18

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 displacement is zero, we have x(0)= x_mcos(F1F1F1S1F1F1F10)=0, which implies cos(F1F1F1S1F1F1F10) = 0. Therefore, F1F1F1S1F1F1F10 is either F1F1F1S1F1F1F10/2 or F1F1F1S1F1F1F10. Now, since the initial velocity is in the negative x direction, we have v(0)= -x_mF1F1F1S1F1F1F10sin(F1F1F1S1F1F1F10) < 0, which implies sin(F1F1F1S1F1F1F10) > 0. Therefore, F1F1F1S1F1F1F10 must be in the first or second quadrant, which means F1F1F1S1F1F1F10/2 < F1F1F1S1F1F1F10 < F1F1F1S1F1F1F10. Therefore, the only possible answer is B.

Question 19

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 answer of A certain spring elongates 9 mm when...

Question 20

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...

Quiz 15: Oscillations

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 0.25-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:

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

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:

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 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:

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:

In simple harmonic motion, the magnitude of the acceleration 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:

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

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

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 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:

In simple harmonic motion:

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?

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:

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:

Quiz 15: Oscillations

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 0.25-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:

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

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:

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 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:

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:

In simple harmonic motion, the magnitude of the acceleration 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:

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

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

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 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:

In simple harmonic motion:

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?

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:

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:

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:

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 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 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: