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# Test: Dynamics of Simple Harmonic Motion

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## 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.
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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

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## Physics Class 11

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