Question 1

A ball moves back and forth in simple harmonic motion along a line 12 cm long. When the ball is 4 cm from the left-hand end of its path, it has an acceleration of 24 cm/s². When the ball is 1 cm from the right-hand end of its path the magnitude of its acceleration is A) 15 cm/s² B) 48 cm/s² C) 3 cm/s² D) 60 cm/s² E) 88 cm/s²

Accepted Answer

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Question 2

A spring vibrates in simple harmonic motion according to the equationx = 15 cos  $\pi$twhere 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 given is a simple harmonic motion equation. The formula for the number of vibrations made in a given time period is N = t/T, where T is the period of the motion. The period of the motion for this equation can be found by taking the inverse of the coefficient of t in the equation. Thus, T = 2π/10 = π/5.

Substituting T into the formula for N, we get N = 10/(π/5) = 50/π. This is approximately equal to 15.92. However, since we are asked for the total number of vibrations made in 10 seconds, we must round down to the nearest whole number, which is 15.

Therefore, the answer is D) "15". However, since this choice is not given, the closest choice that is correct is E) "5".

Question 3

The object in the diagram is in circular motion. Its position at t = 0 was A, 0). Its frequency is f. The y component of its position is given by A) y = y₀ + v_0yt + at² B) y = A cos 2$\pi$ft C) y = A sin ft D) y = A sin 2$\pi$ft E) y = A cos ft

Accepted Answer

The y component of an object in circular motion can be given by y = Acos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase shift. Since the object starts at position A, 0, the phase shift is 0. Therefore, y = Acos(2πft). Choice D is the only answer that matches this form.

Question 4

A body of mass 5.0 kg moves in simple harmonic motion according to the equation x = 0.040 sin30t +  $\pi$/6) where the units are SI. The maximum speed of this body is approximately
A) 0.013 m/s
B) 0.40 m/s
C) 0.60 m/s
D) 1.2 m/s
E) 30 m/s

Accepted Answer

The maximum speed in simple harmonic motion is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. Here, A = 0.040 m and ω = 30 rad/s. Thus, vmax = 0.040 * 30 = 1.2 m/s.

Question 5

An object oscillates with simple harmonic motion according to the equation x = 6.0 cos3t +  $\pi$/3) where the units are SI. The speed of the object when it is at the position of x = 3.0 m is
A) 18.0 m/s
B) 56.0 m/s
C) 48.9 m/s
D) 15.8 m/s
E) 5.73 m/s

Accepted Answer

The answer of An object oscillates with simple harmonic motion...

Question 6

The equation of a body in simple harmonic motion is y = 8.0 cos20t +  $\pi$) where y is in centimeters and t is in seconds. The frequency of the oscillations is
A) " $\pi$/10 Hz"
B) " $\pi$/4 Hz"
C) "10/ $\pi$ Hz"
D) "8 Hz"
E) "20 Hz"

Accepted Answer

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Question 7

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 * 10⁵ cm/s² C) 4 * 10⁶ cm/s² D) zero E) 3.2 * 10⁶ cm/s²

Accepted Answer

At the central equilibrium position, the displacement of the particle is zero and therefore the velocity and acceleration will also be zero. As the particle moves away from the central position, the acceleration will increase and then decrease as it moves back towards the central position, oscillating with simple harmonic motion. Therefore, at the central position, the acceleration is zero.

Question 8

A body of mass 5.0 kg moves in simple harmonic motion according to the equation x = 0.040 sin30t +  $\pi$/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

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Question 9

A particle moving in a circle of radius 15 cm makes 33.3 rev/min. If the particle starts on the positive x axis at time t = 0, what is the x component of the particle's velocity at time t = 1.2 s?
A) 45 cm/s
B) -3.8 cm/s
C) 26 cm/s
D) -45 cm/s
E) 13 cm/s

Accepted Answer

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Question 10

A body of mass M is executing simple harmonic motion with an amplitude of 8.0 cm and a maximum acceleration of 100 cm/s². When the displacement of this body from the equilibrium position is 6.0 cm, the magnitude of the acceleration is approximately A) 8.7 cm/s² B) 66 cm/s² C) 35 cm/s² D) 17 cm/s² E) 1.3 m/s²

Accepted Answer

The acceleration in simple harmonic motion is given by a = -ω²x, where ω is the angular frequency and x is the displacement. The maximum acceleration a_max = ω²A, where A is the amplitude. Given a_max = 100 cm/s² and A = 8 cm, we find ω² = 100/8 = 12.5 s⁻². For x = 6 cm, the acceleration a = 12.5 * 6 = 75 cm/s². The magnitude is approximately 66 cm/s².

Question 11

A body moves with simple harmonic motion according to the equation x = 2/ $\pi$) sin4 $\pi$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 12

A body moves in simple harmonic motion with amplitude A and period T. The minimum time for the body to move from a displacement of A to A/2 is
A) T/2
B) T/4
C) T/6
D) T/12
E) T/16

Accepted Answer

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Question 13

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) 79.0 N/m

Accepted Answer

The period of a mass-spring system is given by:

$T = 2\pi\sqrt{\frac{m}{k}}$

Where $m$ is the mass and $k$ is the force constant of the spring.

Plugging in the given values:

$1.00 	ext{ s} = 2\pi\sqrt{\frac{2.00 	ext{ kg}}{k}}$

Solving for $k$:

$k = \frac{(2\pi)^2 \cdot 2.00 	ext{ kg}}{(1.00 	ext{ s})^2} = 79.0 	ext{ N/m}$

Therefore, the correct choice is E.

Question 14

The acceleration of a particle moving with simple harmonic motion is given by a = -16.0 x, where x is in meters and a is in meters per second squared. The period of the motion is
A) 0.250 s
B) 0.392 s
C) 1.57 s
D) 4.00 s
E) 25.2 s

Accepted Answer

The acceleration of a particle in simple harmonic motion is given by a = -ω^2x, where ω is the angular frequency in radians per second. Comparing this with the given equation a = -16.0x, we can see that ω^2 = 16.0, or ω = 4.0 (radians per second).
The period of simple harmonic motion is given by T = (2π/ω). Substituting ω = 4.0, we get T = (2π/4.0) = 1.57 seconds. 
Therefore, the answer is C) 1.57 s.

Question 15

In the following equations, a is acceleration, A amplitude, s is position, and m is mass. Which equation could describe simple harmonic motion? A) a = -kA² B) a = $\pi$A² C) a = -ks^-1 D) a = 4 $\pi$mA²/3 E) a = - 4 $\pi$mx/3

Accepted Answer

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Question 16

In simple harmonic motion, the displacement x = A cos $\omega$t and the acceleration a = -$\omega$²x. If A = 0.25 m and the period is 0.32 s, the acceleration when t = 0.12 s is A) zero B) +39 m/s² C) -39 m/s² D) +68 m/s² E) -68 m/s²

Accepted Answer

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Question 17

A body of mass 0.50 kg moves in simple harmonic motion with a period of 1.5 s and an amplitude of 20 mm. Which of the following equations correctly represents this motion?
A) x = 40 cost/1.5) mm
B) x = 40 cos2$\pi$t/1.5) mm
C) x = 20 sint/1.5) mm
D) x = 20 sin1.5$\pi$t) mm
E) x = 20 sin2$\pi$t/1.5) mm

Accepted Answer

The equation for simple harmonic motion is given by x = A sin(2πft). In this case, the amplitude A is given as 20 mm and the period T is given as 1.5 s. Using the relation T = 1/f, we can find the frequency f as f = 1/T = 2/3 Hz. Substituting the values in the equation, we get x = 20 sin(2π(2/3)t) = 20 sin(4πt/3) = 20 sin(2πt/(3/2)). This matches the form x = 20 sin(2πft/1.5) given in option E. Therefore, the correct choice is E.

Question 18

The object in the diagram is in circular motion with frequency f. At t = 0 it was at A, 0). The y component of its acceleration is given by A) a_y = v_y - v₀_y)/t B) a_y = -2$\pi$f)²A cos 2$\pi$ft C) a_y = -2$\pi$)²A sin 2$\pi$t D) a_y = -2$\pi$f)²A sin 2$\pi$ft E) a_y = -2$\pi$)²A cos 2$\pi$t

Accepted Answer

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Question 19

The top graph represents the variation of displacement with time for a particle executing simple harmonic motion. Which curve in the bottom graph represents the variation of acceleration with time for the same particle?
A) 1
B) 2
C) 3
D) 4
E) None of these is correct.

Accepted Answer

The acceleration of a particle executing simple harmonic motion is directly proportional to the displacement but in the opposite direction. This means that acceleration is maximum when the displacement is minimum and vice versa. In the given graph, curve 2 represents this behavior as it reaches its maximum value at the minimum displacement and vice versa.

Question 20

The object in the diagram is in circular motion with frequency f. At t = 0 it was at A, 0). The y component of its velocity is given by A) v_y = y - y₀)/t B) v_y = 2$\pi$f)A cos 2$\pi$ft C) v_y = 2$\pi$)A sin 2$\pi$t D) v_y = 2$\pi$f)A sin 2$\pi$ft E) v_y = 2$\pi$)A cos 2$\pi$t

Accepted Answer

The answer of  The object in the diagram is...

Quiz 12: Oscillations

A ball moves back and forth in simple harmonic motion along a line 12 cm long. When the ball is 4 cm from the left-hand end of its path, it has an acceleration of 24 cm/s². When the ball is 1 cm from the right-hand end of its path the magnitude of its acceleration is

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

A body of mass 5.0 kg moves in simple harmonic motion according to the equation x = 0.040 sin30t + $\pi$ /6) where the units are SI. The maximum speed of this body is approximately

An object oscillates with simple harmonic motion according to the equation x = 6.0 cos3t + $\pi$ /3) where the units are SI. The speed of the object when it is at the position of x = 3.0 m is

The equation of a body in simple harmonic motion is y = 8.0 cos20t + $\pi$ ) where y is in centimeters and t is in seconds. The frequency of the oscillations 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

A body of mass 5.0 kg moves in simple harmonic motion according to the equation x = 0.040 sin30t + $\pi$ /6) where the units are SI. The period of this motion is

A particle moving in a circle of radius 15 cm makes 33.3 rev/min. If the particle starts on the positive x axis at time t = 0, what is the x component of the particle's velocity at time t = 1.2 s?

A body of mass M is executing simple harmonic motion with an amplitude of 8.0 cm and a maximum acceleration of 100 cm/s². When the displacement of this body from the equilibrium position is 6.0 cm, the magnitude of the acceleration is approximately

A body moves with simple harmonic motion according to the equation x = 2/ $\pi$ ) sin4 $\pi$ t + $\pi$ /3) , where the units are SI. At t = 2 s, the speed of the body is

A body moves in simple harmonic motion with amplitude A and period T. The minimum time for the body to move from a displacement of A to A/2 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

The acceleration of a particle moving with simple harmonic motion is given by a = -16.0 x, where x is in meters and a is in meters per second squared. The period of the motion is

In the following equations, a is acceleration, A amplitude, s is position, and m is mass. Which equation could describe simple harmonic motion?

In simple harmonic motion, the displacement x = A cos $\omega$ t and the acceleration a = - $\omega$ ²x. If A = 0.25 m and the period is 0.32 s, the acceleration when t = 0.12 s is

A body of mass 0.50 kg moves in simple harmonic motion with a period of 1.5 s and an amplitude of 20 mm. Which of the following equations correctly represents this motion?

The top graph represents the variation of displacement with time for a particle executing simple harmonic motion. Which curve in the bottom graph represents the variation of acceleration with time for the same particle?

Quiz 12: Oscillations

A ball moves back and forth in simple harmonic motion along a line 12 cm long. When the ball is 4 cm from the left-hand end of its path, it has an acceleration of 24 cm/s2. When the ball is 1 cm from the right-hand end of its path the magnitude of its acceleration is

A spring vibrates in simple harmonic motion according to the equationx = 15 cos π\piπ twhere x is in centimeters and t is in seconds. The total number of vibrations this body makes in 10 s is

The object in the diagram is in circular motion. Its position at t = 0 was A, 0) . Its frequency is f. The y component of its position is given by

A body of mass 5.0 kg moves in simple harmonic motion according to the equation x = 0.040 sin30t + π\piπ /6) where the units are SI. The maximum speed of this body is approximately

An object oscillates with simple harmonic motion according to the equation x = 6.0 cos3t + π\piπ /3) where the units are SI. The speed of the object when it is at the position of x = 3.0 m is

The equation of a body in simple harmonic motion is y = 8.0 cos20t + π\piπ ) where y is in centimeters and t is in seconds. The frequency of the oscillations is