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Chapter Test: SHM - 2 - JEE MCQ


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30 Questions MCQ Test - Chapter Test: SHM - 2

Chapter Test: SHM - 2 for JEE 2024 is part of JEE preparation. The Chapter Test: SHM - 2 questions and answers have been prepared according to the JEE exam syllabus.The Chapter Test: SHM - 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Chapter Test: SHM - 2 below.
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Chapter Test: SHM - 2 - Question 1

The acceleration-displacement (a-x) graph of a particle executing simple motion is shown in figure. The frequency of oscillation is given by -


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Chapter Test: SHM - 2 - Question 2

A particle is vibrating in simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position is its energy half potential and half kinetic ?

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Chapter Test: SHM - 2 - Question 3


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Chapter Test: SHM - 2 - Question 4

A pendulum of length 200 cm is horizontal when its bob of mass 200 gm is released. It has 90% of initial energy when the bob reaches the lower most point. Speed of the bob at this point is -

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Chapter Test: SHM - 2 - Question 5

A simple pendulum with angular frequency ω oscillates simple harmonically. The tension in the string at lowest point is T. The total acceleration of the bob at its lowest position is -

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Chapter Test: SHM - 2 - Question 6

What is the spring constant for the combination of spring shown in fig. ?


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Treat the given system as a parallel combination of springs

Chapter Test: SHM - 2 - Question 7

A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. What must be the least period of these oscillation so that the object is not detached ?

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Chapter Test: SHM - 2 - Question 8


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Chapter Test: SHM - 2 - Question 9


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Chapter Test: SHM - 2 - Question 10


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Chapter Test: SHM - 2 - Question 11

A mass m is attached to four springs of spring constants 2k, 2k, k, k as shown in figure. The mass is capable of oscillating on a frictionless horizontal floor. If it is displaced slightly and released the frequency of resulting SHM would be – 


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Chapter Test: SHM - 2 - Question 12

For definite length of wire, if the weight used for applying tension is immersed in water, then frequency will -

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Chapter Test: SHM - 2 - Question 13

Two blocks of masses 10 kg and 2 kg are connected by an ideal spring of spring constant 1000 N/m and the system is placed on a horizontal surface as shown.

The coefficient of friction between 10 kg block and surface is 0.5 but friction is assumed to be absent between 2 kg and surface. Initially blocks are at rest and spring is unstretched then 2 kg block is displaced by 1 cm to elongate the spring then released. Then the graph representing magnitude of frictional force on 10 kg and time t. (time t is measured from that instant when 2kg is released to move)

Detailed Solution for Chapter Test: SHM - 2 - Question 13

Block of mass 2 kg executes S.H.M.

Chapter Test: SHM - 2 - Question 14

A constant horizontal force F is applied on the top of a solid sphere and a hollow sphere of same mass and radius both kept on a sufficiently rough surface. Let a1 and a2 be their linear accelerations, then :


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Chapter Test: SHM - 2 - Question 15


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Chapter Test: SHM - 2 - Question 16


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Chapter Test: SHM - 2 - Question 17

Two particles are in SHM with same angular frequency and amplitudes A and 2A respectively along same straight line with same mean position. They cross each other at position A/2 distance from mean position in opposite direction. The phase difference between them is

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Chapter Test: SHM - 2 - Question 18

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Chapter Test: SHM - 2 - Question 19

If the length of second pendulum is increased by 21%. How many oscillations it will lose per day

Detailed Solution for Chapter Test: SHM - 2 - Question 19

no. of lost oscillations = 43200 – 39273 = 3927

Chapter Test: SHM - 2 - Question 20


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Chapter Test: SHM - 2 - Question 21

A solid ball of mass m is allowed to fall from a height h to a pan suspended with a spring of spring constant k. Assume the ball does not rebound and pan is massless, then amplitude of the oscillation is -


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Chapter Test: SHM - 2 - Question 22


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Chapter Test: SHM - 2 - Question 23

A solid disk of radius R is suspended from a spring of linear spring constant k and torsional constant c, as shown in figure. In terms of k and c, what value of R will give the same period for the vertical and torsional oscillations of this system?


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Chapter Test: SHM - 2 - Question 24

The periodic time of S.H.M. of amplitude 2 cm is 5 sec. If the amplitude is made 4 cm, its periodic time will be -

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Time period is independent with amplitude.

Chapter Test: SHM - 2 - Question 25

The motion of a particle is expressed by the equation acc. a = – bx where x is displacement from equilibrium position and b is constant. What is the periodic time ?

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Chapter Test: SHM - 2 - Question 26

A particle of mass m is executing oscillation about the origin on X-axis. Its potential energy is V(x)=K∣x∣3. Where K is a positive constant. If the amplitude of oscillation is a, then its time period T is proportional to.

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Chapter Test: SHM - 2 - Question 27

 If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes

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SHM is a 1D projection of 2D UCM.

Chapter Test: SHM - 2 - Question 28

 If the sign in equation F = -kx is changed what would happen to the motion of the oscillating body?

Detailed Solution for Chapter Test: SHM - 2 - Question 28

If the sign is changed in F=-kx then the force and hence acceleration will not be opposite to the displacement. Due to this the particle will not oscillate and would accelerate in the direction of displacement. Hence the motion of the body will become linearly accelerated motion.

Chapter Test: SHM - 2 - Question 29

What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×105 N/m and total mechanical energy of 230 J.

Detailed Solution for Chapter Test: SHM - 2 - Question 29

K. Σ=1/2 K(A2-x2)
Max of mean position,
K. Σ=1/2 KA2
=1/2 x4x105x(3x10-2)2
=180J
T.M. Σ=180+P.Σ
230=180+P.Σ
P.Σ=230-180
P.Σ=50J

Chapter Test: SHM - 2 - Question 30

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is

Detailed Solution for Chapter Test: SHM - 2 - Question 30

The relation between angular frequency and displacement is given as
v=ω√A2−x2
Suppose
x=A sinω t
On differentiating the above equation w.r.t. time we get
dx/dt​=Aωcosωt
The maximum value of velocity will be [{v{\max }} = A\omega \]
The displacement for the time when speed is half the maximum is given as
v=Aω/2
A2ω2=4ω(A2−x2)
By substituting the value in (1) we get the displacement as
x=A√3/2

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