Test: Dynamics of SHM


10 Questions MCQ Test Physics For JEE | Test: Dynamics of SHM


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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

Solution:

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

QUESTION: 2

The dimensions and unit of phase constant Φ is

Solution:

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

QUESTION: 3

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

Solution:

SHM is a 1D projection of 2D UCM.

QUESTION: 4

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

Solution:
QUESTION: 5

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

Solution:
QUESTION: 6

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

Solution:

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

QUESTION: 7

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

Solution:

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.

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?

Solution:

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

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.

Solution:

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

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

Solution:

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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