Test: Illustrations of Bodies in Simple Harmonic Motion - JEE MCQ

The reduced mass of the system
M₀=m₁ x m₂/ m₁ + m₂=3x2/3+2=6/5 Kg
Inertia factor=6/5
Spring constant, k=200Nm^-1
Frequency u=1/2π√spring factor/inertia factor
=1/2π√200/6/5
=2.05Hz

The instantaneous displacement of a simple pendulum oscillator is given as y = acos(ωt + π/4)
 
differentiating with respect to time,
dy/dt = -ωasin(ωt + π/4)
here, dy/dt is the velocity of a particle executing SHM.
so, speed of particle = | dy/dt | = ωasin(ωt + π/4)
so, dy/dt will be maximum when sin(ωt + π/4) will be maximum i.e., 1
so, sin(ωt + π/4) = 1 = sin(π/2)
⇒ωt + π/4 = π/2
⇒ωt = π/4
⇒t = π/(4ω)
hence, at t = π/(4ω) , speed of the particle will be maximum.
 

For the pendulum, we can use Newton's second law to write an equation for the forces on the pendulum. The only force responsible for the oscillating motion of the pendulum is the x-component of the weight, so the restoring force on a pendulum is:
F=−mg sinθ
For angles under about 15°, we can approximate sinθ as θ and the restoring force simplifies to:
F ≈ −mgθ
Thus, simple pendulums are simple harmonic oscillators for small displacement angles.

When angle is very small then the value of sin of angle becomes almost 1. This is the one of the necessary conditions for SHM

As the maximum velocities are equal
a1ω1 = a2ω2
a1/a2 = ω2/ω1 ----- (1)
as, k1 = mω1² ----- (2)
k2 = mω2² ----- (3)
substituting eq 2 and eq 3 in eq 1 ,we get
a1/a2 = (k1/k2)^1/2

It is because the total displacement of the string is zero.

The tension in the string would be maximum when bob will be passing mean position.

E α A² and hence the energy will become 9 times.
T is independent of A and hence no change will occur.
Hence C is the correct answer.

The period as usual is given by T = 2n√(m/k). Here `m' is the same as the mass of the bob. The force constant can be found by writing the expression for Young's modulus (since it arises from the elastic force in the cord): V = FL/A(δL) where 6L is the increase in the length of the cord on pulling the bob down with a force F. Therefore, the force constant, F/(δL) = YA/L. On substituting this value, the period is 2m✓(mL/YA).