Simple Harmonic Motion Test

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Simple Harmonic Motion Test - 24

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Question 1
1 / -0

The time period of the following periodic motions respectively are:

(a) sinωt + cosωt

(b) sinωt + cos2ωt + sin4ωt

A

B

C

D

Solution

Hint: The period of a periodic function is the smallest time interval over which the function repeats itself.

Step 1: Find the time period for (sinωt + cosωt).

The periodic time of the function is 2π/ω.

Step 2: Find the time period for (sinωt + cos2ωt + sin4ωt).

The time period is the least interval of time after which a function repeats its value.

The time period of sinωt, T₀= 2π/ω;

The period of the first term is a multiple of the periods of the last two terms.

Therefore, the smallest interval of time after which the sum of the three terms repeats is T₀, and thus, the time period of the sum is 2π/ω.
Question 2
1 / -0

A body oscillates with SHM according to the equation (in SI units) x = 5cos [2π t + π/4].

At t = 1.5 s, the displacement and speed of the body respectively are:

A
3.535 m and 24 m/s

B
−3.535 m and 24 m/s

C
−3.535 m and 22 m/s

D
3.535 m and 22 m/s

Solution

Step 1: Find displacement at t = 1.5 sec.

At t = 1.5 sec,

Displacement = (5.0 m) cos [(2π s^–1) ×1.5 s + π/4]

= (5.0 m) cos [(3π + π/4)]

= –5.0 × 0.707 m

= –3.535 m

Step 2: Find speed at t = 1.5 sec.

Speed of the body = – (5.0 m)(2π s^–1) sin [(2π s^–1) ×1.5 s + π/4]

= – (5.0 m)(2π s^–1) sin [(3π + π/4)]

= 10π × 0.707 m s^–1

= 22 m s^–1
Question 3
1 / -0

What is the length of a simple pendulum, which ticks seconds?

A
0.5 m

B
1 m

C
5 m

D
2 m

Solution

Hint: The time period of a simple pendulum, which ticks seconds, is 2 sec.

Step 1: Find the length of the simple pendulum.

The time period of a simple pendulum is given by,
Question 4
1 / -0

A 5 kg collar is attached to a spring of spring constant 500 N/m. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. The period of oscillation of the collar is:

A
6.3 sec

B
63 sec

C
0.63 sec

D
6 sec

Solution

Hint: The time period of a spring-block system does not depend on its net acceleration.

Step 1: Find the period of oscillation.
Question 5
1 / -0

For the damped oscillator shown in the figure, the mass m of the block is 200 g, k = 90 N/m and the damping constant b is 40 g/s. The period of oscillation is

A
0.03 s

B
0.3 s

C
3 s

D
3.4 s

Solution
Question 6
1 / -0

Which of the following statements is/are correct.

(a) Motion of a kink in a longitudinal spring produced by displacing one end of the spring sideways is transverse and longitudinal, both.

(b) Waves produced in a cylinder containing a liquid by moving its piston back and forth is longitudinal.

(c) Waves produced by a motorboat sailing in water are transverse and longitudinal, both.

(d) Ultrasonic waves in air produced by a vibrating quartz crystal are longitudinal.

A
(a), (b) and (d)

B
(b) and (d)

C
(a), (c) and (d)

D
All are correct

Solution

Hint: Recall types of wave motion.

Step 1: Find the types of wave in each case.

The object which handles shear stress will have a transverse wave, the object which handles volumetric stress will have a longitudinal wave and the object which handles both i.e., shear stress as well as volumetric stress will have both transverse as well as a longitudinal wave

(a) Spring can restore its shape i.e., it can handle the shear stress and volumetric stress so it can have both i.e., transverse as well as a longitudinal wave.

(b) Liquid in a cylinder is not having a free surface so, it can only handle volumetric stress it can not handle shear stress. Hence, only a longitudinal wave is possible in this case.

(c) Waves on the water surface if both i.e., transverse as well as longitudinal

(d) As air being a fluid so it can only handle the volumetric stress so, in this case only longitudinal wave is possible.
Question 7
1 / -0

A body oscillates with SHM according to the equation (in SI units),
x = 5 cos [2π t + π/4].
At t = 1.5 s, acceleration of the body will be

A
140 cm /s²

B
160 m /s²

C
140 m /s²

D
14 m/s²

Solution
Question 8
1 / -0

A body is executing simple harmonic motion with frequency 'n', the frequency of its potential energy is:

A
3n

B
4n

C
n

D
2n

Solution

Hint: The potential energy changes two times in one oscillation.

As the potential energy changes two times in an oscillation, the frequency of the potential energy is 2n.
Question 9
1 / -0

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is:

A
3.14 s

B
0.628 s

C
0.0628 s

D
6.28 s

Solution
Question 10
1 / -0

Which one of the following statement/s is correct about simple harmonic motion?

A
Simple harmonic motion can take place in a noninertial frame

B
In a noninertial frame the ratio of the force applied with the displacement should be constant

C
Simple harmonic motion can not take place in a noninertial frame

D
Both (2) and (3)

Solution

Consider an accelerating vehicle in which a block is attached to a spring on a smooth horizontal surface and the other end is fixed. The rest position of the block will change compared to the nonaccelerating vehicle. If the block is stretched and released it will be in a simple harmonic motion having the mean position changed to a new one. But the ratio of the applied force with the displacement will not be constant until we take a pseudo force into account that is mass times the acceleration of the vehicle and opposite to the direction of the accelerating frame.