NEET  >  Physics Class 11  >  Test: Dynamics of Simple Harmonic Motion Download as PDF

Test: Dynamics of Simple Harmonic Motion


Test Description

10 Questions MCQ Test Physics Class 11 | Test: Dynamics of Simple Harmonic Motion

Test: Dynamics of Simple Harmonic Motion for NEET 2023 is part of Physics Class 11 preparation. The Test: Dynamics of Simple Harmonic Motion questions and answers have been prepared according to the NEET exam syllabus.The Test: Dynamics of Simple Harmonic Motion MCQs are made for NEET 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Dynamics of Simple Harmonic Motion below.
Solutions of Test: Dynamics of Simple Harmonic Motion questions in English are available as part of our Physics Class 11 for NEET & Test: Dynamics of Simple Harmonic Motion solutions in Hindi for Physics Class 11 course. Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free. Attempt Test: Dynamics of Simple Harmonic Motion | 10 questions in 10 minutes | Mock test for NEET preparation | Free important questions MCQ to study Physics Class 11 for NEET Exam | Download free PDF with solutions
1 Crore+ students have signed up on EduRev. Have you?
Test: Dynamics of Simple Harmonic Motion - Question 1

Potential energy of a particle with mass m is U=k[x]3 , where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period, T is

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 1

U=Kx3
F=-du/ 
F=-Kx ------(2)
F=-3kx2 -----(1)
Comparing equation 1 and 2
K=-3kx
T=2π√m/K
Because, x=Asinwt x∝A
T∝1/√k ∝1/√A

Test: Dynamics of Simple Harmonic Motion - Question 2

The dimensions and unit of phase constant Φ is

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 2

Mathematical constants don't have any dimensions hence, it is dimensionless and phase constant is in angle therefore, the unit is radian.

Test: Dynamics of Simple Harmonic Motion - Question 3

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

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 3

SHM is a 1D projection of 2D UCM.

Test: Dynamics of Simple Harmonic Motion - Question 4

The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is

Test: Dynamics of Simple Harmonic Motion - Question 5

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

Test: Dynamics of Simple Harmonic Motion - Question 6

Choose the correct time period of the function sin ωt + cos ωt

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 6

If the time period of f(x) = T
then time period of f(ax+b) = aT
the time period of sint+cost= 2π
so, time period of sinωt+cosωt = 2π/ω

Test: Dynamics of Simple Harmonic Motion - Question 7

The velocity and acceleration amplitudes of body executing simple harmonic motion is

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 7

Maximum velocity: v = ωA, where ω is uniform angular velocity and a is the radius of the circle in which a reference particle executing S.H.M.
Velocity is maximum at mean positions. The maximum value of velocity is called velocity amplitude in SHM.
Acceleration is maximum at extreme position given by A = - ω2A. The maximum value of acceleration is called acceleration amplitude in SHM.

Test: Dynamics of Simple Harmonic Motion - Question 8

At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 8

Let say from some distance x, the KE = PE and as total energy must be conserved and TE = -½ kA2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2

Test: Dynamics of Simple Harmonic Motion - Question 9

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 Test: Dynamics of Simple Harmonic Motion - Question 9

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

Test: Dynamics of Simple Harmonic Motion - Question 10

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 Test: Dynamics of Simple Harmonic Motion - Question 10

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

127 videos|464 docs|210 tests
Use Code STAYHOME200 and get INR 200 additional OFF
Use Coupon Code
Information about Test: Dynamics of Simple Harmonic Motion Page
In this test you can find the Exam questions for Test: Dynamics of Simple Harmonic Motion solved & explained in the simplest way possible. Besides giving Questions and answers for Test: Dynamics of Simple Harmonic Motion , EduRev gives you an ample number of Online tests for practice