JEE Exam  >  JEE Tests  >  Physics for JEE Main & Advanced  >  Chapter Test: SHM - 2 - JEE MCQ

Chapter Test: SHM - 2 - JEE MCQ


Test Description

30 Questions MCQ Test Physics for JEE Main & Advanced - Chapter Test: SHM - 2

Chapter Test: SHM - 2 for JEE 2024 is part of Physics for JEE Main & Advanced 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.
Solutions of Chapter Test: SHM - 2 questions in English are available as part of our Physics for JEE Main & Advanced for JEE & Chapter Test: SHM - 2 solutions in Hindi for Physics for JEE Main & Advanced course. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free. Attempt Chapter Test: SHM - 2 | 30 questions in 60 minutes | Mock test for JEE preparation | Free important questions MCQ to study Physics for JEE Main & Advanced for JEE Exam | Download free PDF with solutions
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 -


Detailed Solution for Chapter Test: SHM - 2 - Question 1


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 ?

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


1 Crore+ students have signed up on EduRev. Have you? Download the App
Chapter Test: SHM - 2 - Question 3


Detailed Solution for Chapter Test: SHM - 2 - Question 3


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 -

Detailed Solution for Chapter Test: SHM - 2 - Question 4


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 -

Detailed Solution for Chapter Test: SHM - 2 - Question 5


Chapter Test: SHM - 2 - Question 6

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


Detailed Solution for Chapter Test: SHM - 2 - Question 6

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 ?

Detailed Solution for Chapter Test: SHM - 2 - Question 7


Chapter Test: SHM - 2 - Question 8


Detailed Solution for Chapter Test: SHM - 2 - Question 8


Chapter Test: SHM - 2 - Question 9


Detailed Solution for Chapter Test: SHM - 2 - Question 9


Chapter Test: SHM - 2 - Question 10


Detailed Solution for Chapter Test: SHM - 2 - Question 10


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 – 


Detailed Solution for Chapter Test: SHM - 2 - Question 11


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 -

Detailed Solution for Chapter Test: SHM - 2 - Question 12


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 :


Detailed Solution for Chapter Test: SHM - 2 - Question 14


Chapter Test: SHM - 2 - Question 15


Detailed Solution for Chapter Test: SHM - 2 - Question 15

Chapter Test: SHM - 2 - Question 16


Detailed Solution for Chapter Test: SHM - 2 - Question 16


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

Detailed Solution for Chapter Test: SHM - 2 - Question 17


Chapter Test: SHM - 2 - Question 18

Detailed Solution for Chapter Test: SHM - 2 - Question 18


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


Detailed Solution for Chapter Test: SHM - 2 - Question 20


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 -


Detailed Solution for Chapter Test: SHM - 2 - Question 21


Chapter Test: SHM - 2 - Question 22


Detailed Solution for Chapter Test: SHM - 2 - Question 22


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?


Detailed Solution for Chapter Test: SHM - 2 - Question 23


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 -

Detailed Solution for Chapter Test: SHM - 2 - Question 24

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 ?

Detailed Solution for Chapter Test: SHM - 2 - Question 25


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.

Detailed Solution for Chapter Test: SHM - 2 - Question 26

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

Detailed Solution for Chapter Test: SHM - 2 - Question 27

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

289 videos|635 docs|179 tests
Information about Chapter Test: SHM - 2 Page
In this test you can find the Exam questions for Chapter Test: SHM - 2 solved & explained in the simplest way possible. Besides giving Questions and answers for Chapter Test: SHM - 2, EduRev gives you an ample number of Online tests for practice

Top Courses for JEE

289 videos|635 docs|179 tests
Download as PDF

Top Courses for JEE