Question 1

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

Accepted Answer

Let's use the equation of motion for a particle in simple harmonic motion: x(t) = A*cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle. We know that the period T = 1.5s, so ω = (2π/T) = (4π/3) rad/s. At t₀ = 0, the particle is at the equilibrium point, which means that x(0) = A*cos(φ) = 0. We also know that the velocity at this point is 1.00 m/s to the right, so v(0) = -A*ω*sin(φ) = 1.00 m/s. Solving for A and φ, we get A = 1.00/(ω*sin(φ)) and φ = arcsin(-v(0)/(A*ω)). At time t later, the particle has moved to the left with a velocity of 0.50 m/s, which means that v(t) = -A*ω*sin(ωt + φ) = -0.50 m/s. Substituting the expressions for A and φ into the equation for v(t), we get: sin(ωt + φ) = -0.50/(1.00/ω) Using the identity sin(-θ) = -sin(θ), we can rewrite this as: sin(ωt - |φ|) = 0.50/(1.00/ω) The smallest possible value of t occurs when the sine function equals 1. Note that sin(θ) can only equal 1 when θ = π/2 or θ = 3π/2. So we have two possible values for ωt - |φ|: π/2 or 3π/2. For each value, we can solve for t: ωt - |φ| = π/2 t = (π/2 + |φ|)/ω ωt - |φ| = 3π/2 t = (3π/2 + |φ|)/ω Taking the smaller of these two solutions, we get: t = (π/2 + |φ|)/ω Substituting the expressions for A and φ into this formula, we get: t = (π/2 + arcsin(v(0)/(A*ω)))/ω Plugging in the values we know, we get: t = (π/2 + arcsin(1.00/(1.00*sin(φ))))/ω t = (π/2 + arcsin(1.00/(1.00*sin(arcsin(-v(0)/(A*ω))))))/ω t = (π/2 + arcsin(1.00/(1.00*sin(arcsin(-1.00/(1.00*sin(φ)))))))/(4π/3) Simplifying this expression using the identity arcsin(sin(θ)) = θ, we get: t = (π/2 + arcsin(-1.00/sin(φ)))/(4π/3) t = (3/8 - arcsin(1.00/sin(φ)))/(2π/3) Using the identity sin(-θ) = -sin(θ) and the fact that φ is in the first quadrant, we can simplify the expression for t to: t = (3/8 - arcsin(1.00/A))/(2π/3) We want to minimize t, so we need to maximize arcsin(1.00/A). The maximum value of arcsin(1.00/A) occurs when A is minimized, so we need to find the minimum value of A. Expressing A in terms of v(0) and ω, we get: A = |v(0)|/ω A = 1.00/(ω*sin(arcsin(-v(0)/(A*ω)))) A = 1.00/sqrt(1 - v(0)^2/ω^2) Plugging in the known values, we get: A = 2.00 m Substituting this value of A into the expression for t, we get: t = (3/8 - arcsin(0.50/2.00))/(2π/3) t = (3/8 - 0.2527)/(2π/3) t = 0.50 s Therefore, the smallest possible value of t is 0.50 s, which corresponds to choice (C).

Question 2

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

Accepted Answer

D

Question 3

Graph shows position of an oscillating particle as a function of time. What is the amplitude of the oscillation?

Accepted Answer

The amplitude of the oscillation is the maximum displacement from the equilibrium position, which is half the distance between the highest and lowest points on the graph. In this case, the highest point is 4.0 cm and the lowest point is -4.0 cm, so the amplitude is (4.0 cm - (-4.0 cm))/2 = 4.0 cm.

Question 4

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

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The answer of A particle moving with a simple harmonic...

Question 5

Restoring force is such that it always pushes or pulls an object to its maximum displacement.

Accepted Answer

Restoring force pushes or pulls an object towards its equilibrium position, not its maximum displacement.

Question 6

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

Accepted Answer

The answer of The frequency of a simple harmonic motion...

Question 7

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

Accepted Answer

The instantaneous speed of a mass undergoing simple harmonic motion is given by the formula v = Aωcos(ωt), where A is the amplitude of oscillation, ω is the angular frequency (which is related to the frequency by ω = 2πf), and t is the time at which the speed is measured. Therefore, the instantaneous speed of the mass depends on the amplitude, frequency, and the time at which the speed is measured. Additionally, the period of oscillation (which is the inverse of the frequency) is also a factor in determining the speed, since it affects how quickly the mass completes one full cycle of motion. Therefore, all of the given options contribute to determining the instantaneous speed of a mass undergoing simple harmonic motion.

Question 8

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

Accepted Answer

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

Question 9

A spring is cut in half. The ratio of the force constant of one of the halves to the force constant of the original spring is

Accepted Answer

The answer of A spring is cut in half. The...

Question 10

A particle with a mass of 65 g is undergoing 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

Accepted Answer

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

Question 11

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

Accepted Answer

The period of simple harmonic motion for a mass hanging on a spring is given by:

T = 2π√(m/k)

When the same mass is hung from a spring with a spring constant of 2k, the period of oscillation can be found by substituting 2k for k in the above equation:

T' = 2π√(m/2k) = (1/√2) * 2π√(m/k) = (1/√2) * T

Therefore, the period of oscillation decreases by a factor of approximately 1.41 or √2. The closest option is D, which states that the period decreases by a factor of approximately 1.41.

Question 12

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

Accepted Answer

The period of the motion can be found using the time it takes for the particle to move from maximum positive displacement to maximum negative displacement:
T = 2.25 s

The amplitude of the motion is the maximum displacement, which is 12.0 cm. Therefore, the equation for the motion is:
x = 12.0 cos(2πt/T)

We need to find the time it takes for the particle to move from x = +12.0 cm to x = -6.00 cm. Let's set up the equation:
-6.00 = 12.0 cos(2πt/T)

Divide both sides by 12.0:
-0.500 = cos(2πt/T)

Take the inverse cosine of both sides:
2πt/T = cos⁻¹(-0.500) = 2.094

Solve for t:
t = (2.094 T) / (2π) = 1.50 s

Therefore, the time for the particle to move to x = -6.00 cm is 1.50 s, which corresponds to choice C.

Question 13

Newborn baby's heart beats 120 times per minute. What is the frequency of baby's heartbeat?

Accepted Answer

The frequency of the baby's heartbeat can be calculated as the number of beats per unit of time. In this case, the baby's heart beats 120 times per minute, so the frequency is 120/60 = 2.0 Hz.

Question 14

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

Accepted Answer

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

Question 15

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

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 16

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

Accepted Answer

In simple harmonic motion, the force is directly proportional to the displacement and acts in the opposite direction, which is represented by F = -kx.

Question 17

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

Accepted Answer

In simple harmonic motion, the maximum speed of the object occurs at the equilibrium point, where the net force acting on the object is zero. This is because, at this point, the potential energy of the object is at its minimum and the kinetic energy is at its maximum. At the highest and lowest points, the object briefly comes to rest before changing direction, so its speed is momentarily zero at these points.

Question 18

It takes an oscillating particle 3.0 s to travel from maximum displacement to equilibrium position. What is the period of the oscillation?

Accepted Answer

The answer of It takes an oscillating particle 3.0 s...

Question 19

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

Accepted Answer

The answer of A mass m hanging on a spring...

Question 20

In order for oscillations to occur, the restoring force has to be proportional to an object's displacement from equilibrium.

Accepted Answer

The answer of In order for oscillations to occur, the...

Quiz 12: Oscillations

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

Graph shows position of an oscillating particle as a function of time. What is the amplitude of the oscillation?

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

Restoring force is such that it always pushes or pulls an object to its maximum displacement.

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

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

A spring is cut in half. The ratio of the force constant of one of the halves to the force constant of the original spring is

A particle with a mass of 65 g is undergoing 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 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 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

Newborn baby's heart beats 120 times per minute. What is the frequency of baby's heartbeat?

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

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

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

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

It takes an oscillating particle 3.0 s to travel from maximum displacement to equilibrium position. What is the period of the oscillation?

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

In order for oscillations to occur, the restoring force has to be proportional to an object's displacement from equilibrium.

Quiz 12: Oscillations

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

Graph shows position of an oscillating particle as a function of time. What is the amplitude of the oscillation?

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

Restoring force is such that it always pushes or pulls an object to its maximum displacement.

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

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

A spring is cut in half. The ratio of the force constant of one of the halves to the force constant of the original spring is

A particle with a mass of 65 g is undergoing 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 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 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

Newborn baby's heart beats 120 times per minute. What is the frequency of baby's heartbeat?

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

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

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

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

It takes an oscillating particle 3.0 s to travel from maximum displacement to equilibrium position. What is the period of the oscillation?

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

In order for oscillations to occur, the restoring force has to be proportional to an object's displacement from equilibrium.

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

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