Test: Dynamics of Simple Harmonic Motion - NEET MCQ

Test: Dynamics of Simple Harmonic Motion - NEET MCQ

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Test: Dynamics of Simple Harmonic Motion - Question 1

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 Test: Dynamics of Simple Harmonic Motion - Question 1

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.

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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

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 4

Maximum acceleration=ω2r
= (0.45)2x3
=0.60ms-2

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?

Detailed Solution for Test: Dynamics of Simple Harmonic Motion - Question 5

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.

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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