Question 1

A particle moving with simple harmonic motion has a maximum displacement of
+12)0 cm. The particle moves from its maximum positive to its maximum negative displacement in 2.25 s. The motion starts when the displacement is x = +12.0 cm. The time for the particle to move to x = -6.00 cm is
A) 1.70 s
B) 0.750 s
C) 1.50 s
D) 2.20 s
E) 0.983 s

Accepted Answer

The motion of the particle can be described by the following equation:

x = A cos(ωt + φ)

where x is the displacement from the equilibrium position, A is the amplitude (maximum displacement), ω is the angular frequency, t is time, and φ is the phase angle.

From the problem statement, we know that the maximum displacement is +12.0 cm, so A = 12.0 cm. We also know that the particle takes 2.25 s to go from its maximum positive displacement to its maximum negative displacement, which means that half the period, T/2, is 2.25 s.

The period of simple harmonic motion is given by:

T = 2π / ω

Since we know that T/2 = 2.25 s, we can solve for ω:

T/2 = π / ω
ω = π / (T/2)
ω = 2π / T

Substituting T/2 = 2.25 s, we get:

ω = 2π / (2 × 2.25)
ω = 1.40 rad/s

Now we can use the equation for x to find the time it takes for the particle to move to x = -6.00 cm:

-6.00 = 12.0 cos(1.40t + φ)

Solving for the phase angle, we get:

φ = acos(-6.00/12.0)
φ = 2.09 rad

Substituting A = 12.0 cm, ω = 1.40 rad/s, and φ = 2.09 rad, we get:

-6.00 = 12.0 cos(1.40t + 2.09)

We can solve for t using algebra or a graphing calculator. The answer is t = 1.50 s. Therefore, the best choice is C.

Question 2

A spring vibrates in simple harmonic motion according to the equation
X = 15 cos $\pi$t. 
Where x is in centimeters and t is in seconds. The total number of vibrations this body makes in 10 s is
A) 0.5
B) 10
C) $\pi$
D) 15
E) 5

Accepted Answer

The equation is in the form X = A cos(ωt), where ω is the angular frequency. Here, ω = F1F1F1S1F1F1F10. The frequency f is ω/2π = F1F1F1S1F1F1F10/2π = 5 Hz. In 10 seconds, the number of vibrations is 5 Hz * 10 s = 50 vibrations.

Question 3

A particle moving with a simple harmonic motion has its maximum displacement of +18 cm at time t = 0. The frequency of the motion is 10 s^-1. At a time t = 0.65 s, the position of the particle is A) +18 cm B) zero C) -13 cm D) -18 cm E) +7.3 cm

Accepted Answer

D

Question 4

A body moves with simple harmonic motion according to the equation
X = (2/$\pi$) sin(4$\pi$t + $\pi$/3),
Where the units are SI. At t = 2 s, the speed of the body is
A) 1/3 m/s
B) 1/$\pi$ m/s
C)  m/s
D) 4 m/s
E)  m/s

Accepted Answer

The answer of A body moves with simple harmonic motion...

Question 5

A body oscillates with simple harmonic motion according to the equation
X = 6.0 cos(3t + $\pi$/3)
Where the units are SI. The speed of the body when it has a displacement of
3 m is
A)  m/s
B) 6$\pi$m/s
C) 9$\pi$ m/s
D)  m/s
E) 18$\pi$ m/s

Accepted Answer

We can find the velocity of the body by taking the derivative of the displacement equation:

v = dx/dt = -18 sin(3t + π/3)

To find the velocity when the body has a displacement of 3 m, we need to plug in X = 3 to this equation and solve for v:

3 = 6 cos(3t + π/3)
cos(3t + π/3) = 1/2
3t + π/3 = ±π/3 + 2nπ   (where n is an integer)
t = (±π/3 - π/3 + 2nπ)/3
t = (±π/9 + nπ/3)

Plugging these values of t into the velocity equation, we find:

v = -18 sin(±π/3) = ±9√3 m/s

There are two possible values for v because there are two values of t that correspond to X = 3 (when the body is at maximum displacement, and when it is returning to that position).

However, we are only asked for the speed (magnitude of velocity), which is the same for both values:

|v| = 9√3 m/s

This matches choice D.

Question 6

A particle with a mass of 65 g is moving with simple harmonic motion. At time t = 0, the particle is at its extreme positive displacement of 18.0 cm. The period of the motion is 0.600 s. At time t = 1.35 s, the velocity of the particle is
A) -1.9 m/s
B) zero
C) 0.84 m/s
D) +1.9 m/s
E) -0.84 m/s

Accepted Answer

The answer of A particle with a mass of 65...

Question 7

The frequency of a simple harmonic motion is 2.6 $\times$ 10^-4 s^-1. The oscillation starts (t = 0) when the displacement has its maximum positive value of 6.5 $\times$ 10^-3 cm. The earliest possible time at which the particle can be found at x = -2.6 $\times$ 10^-3 cm is A) 7.1 $\times$ 10^-6 s B) 1.2 $\times$ 10^-5 s C) 1.1 $\times$ 10^-4 s D) 1.1 $\times$ 10^-3 s E) 4.2 $\times$ 10^-3 s

Accepted Answer

The frequency of the simple harmonic motion is given as $2.6 \times 10^{-4} \, \text{s}^{-1}$, which means the period $T = \frac{1}{f} = \frac{1}{2.6 \times 10^{-4}} \, \text{s}$. The particle starts at its maximum positive displacement, and to reach a displacement of $-2.6 \times 10^{-3} \, \text{cm}$, it needs to pass through the equilibrium position and reach the negative side. The earliest time this can happen is a quarter of the period later, because in simple harmonic motion, the object goes from one extreme, through the equilibrium position, to the other extreme, and back in one period. Calculating a quarter of the period gives the time it takes to reach the specified displacement for the first time.

Question 8

A mass m hanging on a spring with a spring constant k has simple harmonic motion with a period T. If the mass is doubled to 2m, the period of oscillation
A) increases by a factor of 2.
B) decreases by a factor of 2.
C) increases by a factor of 
D) decreases by a factor of 
E) is not affected.

Accepted Answer

The period of oscillation of a spring-mass system is given by T=2π√(m/k). 
If the mass is doubled (m→2m), the period becomes T'=2π√(2m/k)=√2 T.
Therefore, the period increases by a factor √2, which is approximately 1.41, or 11eaa4be_8520_eb9a_9e4a_4b5165ed53c4_TB6080_11. So, the correct answer is C.

Question 9

Any body moving with simple harmonic motion is being acted on by a force that is
A) constant.
B) proportional to a sine or cosine function of the displacement.
C) proportional to the inverse square of the displacement.
D) directly proportional to the displacement.
E) proportional to the square of the displacement.

Accepted Answer

In simple harmonic motion, the restoring force is directly proportional to the displacement from the equilibrium position but acts in the opposite direction. This relationship is described by Hooke's Law, which is fundamental to understanding simple harmonic motion.

Question 10

A 2.00-kg body is attached to a spring of negligible mass and oscillates with a period of 1.00 s. The force constant of the spring is
A) 0.051 N/m
B) 0.500 N/m
C) 2.00 N/m
D) 6.28 N/m
E) None of these is correct.

Accepted Answer

The answer of A 2.00-kg body is attached to a...

Question 11

The equation for the period T of a mass m oscillating with simple harmonic motion at the end of a spring with a force constant k is  . A mass m that is oscillating on a spring with a force constant of 0.52 N/m has a period of 2.1 s. On a second spring, the same mass has a period of 3.5 s. The force constant of the second spring is
A) impossible to determine because the mass is not given.
B) 0.19 N/m
C) 1.4 N/m
D) 0.31 N/m
E) 0.75 N/m

Accepted Answer

The answer of The equation for the period T of...

Question 12

A particle is oscillating with simple harmonic motion. The frequency of the motion is 10 Hz and the amplitude of the motion is 5.0 cm. As the particle passes its central equilibrium position, the acceleration of the particle is A) 100 cm/s² B) 1.6 $\times$ 10⁵ cm/s² C) 4 $\times$ 10⁶ cm/s² D) zero E) 3.2 $\times$ 10⁶ cm/s²

Accepted Answer

At the central equilibrium position, the displacement of the particle is zero, so the velocity is zero too. As the acceleration is proportional to the displacement, the acceleration must also be zero at this point. Therefore, the correct answer is D: zero.

Question 13

The instantaneous speed of a mass undergoing simple harmonic motion on the end of a spring depends on
A) the amplitude of oscillation.
B) the frequency of oscillation.
C) the period of oscillation.
D) the time at which the speed is measured.
E) all of these

Accepted Answer

The instantaneous speed of a mass undergoing simple harmonic motion on the end of a spring depends on all of the given factors.

A) The amplitude of oscillation affects the maximum speed that the mass will achieve during its motion.

B) The frequency of oscillation affects the rate at which the mass moves back and forth, meaning it will affect the speed of the mass at any given point in time.

C) The period of oscillation is related to the frequency and also affects the rate at which the mass moves back and forth, meaning it will affect the speed of the mass at any given point in time.

D) The time at which the speed is measured will affect the instantaneous speed, as the mass will be at a different point in its oscillation at each point in time.

Question 14

When an object is oscillating in simple harmonic motion in the vertical direction, its maximum speed occurs when the object
A) is at its highest point.
B) is at its lowest point.
C) is at the equilibrium point.
D) has the maximum net force exerted on it.
E) has a position equal to its amplitude.

Accepted Answer

At the equilibrium point, the object has zero potential energy and maximum kinetic energy due to the displacement from equilibrium. As it moves away from the equilibrium point, its potential energy increases and kinetic energy decreases, resulting in a decrease in speed. At the highest and lowest points, the object briefly comes to a stop before changing direction, so its speed is momentarily zero. Hence, the maximum speed occurs at the equilibrium point.

Question 15

You want a mass that, when hung on the end of a spring, oscillates with a period of 1 s. If the spring has a spring constant of 10 N/m, the mass should be A) 10 kg B) kg C) 4$\pi$²(10) kg D) 10/(4$\pi$²) kg E) None of these is correct.

Accepted Answer

The answer of You want a mass that, when hung...

Question 16

The force constant of a massless spring is 25.0 N/m. A mass of 0.45 kg is oscillating in simple harmonic motion at the end of the spring with an amplitude of 0.32 m. The maximum speed of the mass is
A) 5.7 m/s
B) 56 m/s
C) 7.4 m/s
D) 2.4 m/s
E) 10 m/s

Accepted Answer

The answer of The force constant of a massless spring...

Question 17

A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t₀ = 0 with a velocity of 1.00 m/s to the right. A time t later, the particle is observed to move to the left with a velocity of 0.50 m/s. (Note the change in direction of the velocity.) The smallest possible value of the time t is A) 0.17 s B) 0.33 s C) 0.50 s D) 0.25 s E) 0.82 s

Accepted Answer

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

Question 18

If F is the force, x the displacement, and k a particular constant, for simple harmonic motion we must have A) F = -k/x² B) F = k/x C) F = (k/x²)^1/2 D) F = -kx² E) None of these is correct.

Accepted Answer

The answer of If F is the force, x the...

Question 19

A body of mass 5.0 kg moves in simple harmonic motion according to the equation 
X = 0.040 sin(30t + $\pi$/6) 
Where the units are SI. The period of this motion is
A) 1/30 s
B) $\pi$/15 s
C) $\pi$/6 s
D) 15/$\pi$ s
E) 30 s

Accepted Answer

The answer of A body of mass 5.0 kg moves...

Question 20

A mass m hanging on a spring with a spring constant k executes simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 2k, the period of oscillation
A) increases by a factor of 2.
B) decreases by a factor of 2.
C) increases by a factor of  .
D) decreases by a factor of  .
E) is not affected.

Accepted Answer

The period of simple harmonic motion is given by T=2π√(m/k), where m is the mass and k is the spring constant. Thus, when the mass is hung from a spring with spring constant 2k, the period becomes T=2π√(m/(2k))=π√(m/k). This is a factor of √2 smaller than the original period, so the period decreases by a factor of √2, which is approximately 1.41 or 11eaa4be_8521_d5fd_9e4a_1712379399fc_TB6080_11. Therefore, the correct choice is D.

Quiz 14: Oscillations

A particle moving with simple harmonic motion has a maximum displacement of +12) 0 cm. The particle moves from its maximum positive to its maximum negative displacement in 2.25 s. The motion starts when the displacement is x = +12.0 cm. The time for the particle to move to x = -6.00 cm is

A spring vibrates in simple harmonic motion according to the equation X = 15 cos $\pi$ t. Where x is in centimeters and t is in seconds. The total number of vibrations this body makes in 10 s is

A particle moving with a simple harmonic motion has its maximum displacement of +18 cm at time t = 0. The frequency of the motion is 10 s^-1. At a time t = 0.65 s, the position of the particle is

A body moves with simple harmonic motion according to the equation X = (2/ $\pi$ ) sin(4 $\pi$ t + $\pi$ /3) , Where the units are SI. At t = 2 s, the speed of the body is

A body oscillates with simple harmonic motion according to the equation X = 6.0 cos(3t + $\pi$ /3) Where the units are SI. The speed of the body when it has a displacement of 3 m is

A particle with a mass of 65 g is moving with simple harmonic motion. At time t = 0, the particle is at its extreme positive displacement of 18.0 cm. The period of the motion is 0.600 s. At time t = 1.35 s, the velocity of the particle is

The frequency of a simple harmonic motion is 2.6 $\times$ 10^-4 s^-1. The oscillation starts (t = 0) when the displacement has its maximum positive value of 6.5 $\times$ 10^-3 cm. The earliest possible time at which the particle can be found at x = -2.6 $\times$ 10^-3 cm is

A mass m hanging on a spring with a spring constant k has simple harmonic motion with a period T. If the mass is doubled to 2m, the period of oscillation

Any body moving with simple harmonic motion is being acted on by a force that is

A 2.00-kg body is attached to a spring of negligible mass and oscillates with a period of 1.00 s. The force constant of the spring is

A particle is oscillating with simple harmonic motion. The frequency of the motion is 10 Hz and the amplitude of the motion is 5.0 cm. As the particle passes its central equilibrium position, the acceleration of the particle is

The instantaneous speed of a mass undergoing simple harmonic motion on the end of a spring depends on

When an object is oscillating in simple harmonic motion in the vertical direction, its maximum speed occurs when the object

You want a mass that, when hung on the end of a spring, oscillates with a period of 1 s. If the spring has a spring constant of 10 N/m, the mass should be

The force constant of a massless spring is 25.0 N/m. A mass of 0.45 kg is oscillating in simple harmonic motion at the end of the spring with an amplitude of 0.32 m. The maximum speed of the mass is

If F is the force, x the displacement, and k a particular constant, for simple harmonic motion we must have

A body of mass 5.0 kg moves in simple harmonic motion according to the equation X = 0.040 sin(30t + $\pi$ /6) Where the units are SI. The period of this motion is

A mass m hanging on a spring with a spring constant k executes simple harmonic motion with a period T. If the same mass is hung from a spring with a spring constant of 2k, the period of oscillation