Simple Harmonic Motion NAT Level - 2 - IIT JAM MCQ

The correct answer is: 8

At this point, the rebound speed of block  1  can be found by:

Parallel axis theorem

d → distance from pivot to centre of mass.
Period of physical pendulum

= 1.2 seconds
The correct answer is: 1.2

I = moment of inertia 
r = moment arm

(for small angles)

∴ 
For the first system
l₁ = 2mr²
radius of gyration = r₁, = 2r 
For the second system

radius of gyration 

The correct answer is: 1.09

Parallel
Effective spring constant = k₁+ k₂ = k_p
Series
Effective spring constant  

The correct answer is: 0.5

  where → amplitude of oscillator 
V_m = 5 cm/s since V_s= 4 cm/s

from the graph, at t = 0, v = 4 cm/s

If we take ,acceleration would be positive but from the graph since the v is decreasing at
t = 0
∴ the acceleration should be negative!
∴ Ø = 53°
The correct answer is: 53

The restoring torque on the bob of the pendulum at an angle with the vertical is given by

We apply small angle approximation to so that the motion can be followed as simple harmonic motion, the torque becomes

The problem requires to find the value of 9 for which

Write the taylor expansion of Sin θ

Taking absolute value on both sides

The correct answer is: 19.65

For the simple harmonic oscillator (SHO) part,

Since the graph shows only half of the usual SHO potential, the period contribution from the SHO part should be half the usual period

For the gravitational potential, one can calculate the period from the usual kinematics equation for constant acceleration

Since E is always conserved

At the end point  
x = Elm g

Since the particle has to travel from the origin to the right end point & then back again the total time contribution is twice

The correct answer is: 2

Assuming both the loops to be point masses concentrated at the centre of mass, they can be considered as simple pendulums. For small oscillations, time period of a
simple pendulum 

Time period of Y is 
The correct answer is: 0.5