Simple Harmonic Motion NAT Level - 1 - IIT JAM MCQ

When spring is cut in equal part, then force constant is 

The correct answer is: 0.707

2n
The correct answer is: 2

Kinetic energy  
Kinetic energy  
The correct answer is: 0.75

The variation in amplitude of a damped harmonic oscillation with time is given by

 initial amplitude,
 damped factor
It is given that after 1 minutes  

The correct answer is: 8

Using series and parallel combination of spring

The correct answer is: 1.5

Potential energy = Kinetic energy

The correct answer is: 2.828

The correct answer is: 0.667

Total energy  
Potential energy 

So total energy 
The correct answer is: 10

The correct answer is: 2

The formula for the time period of a mass-spring system is:
T = 2π √(m/k)
where T is the time period, m is the mass, and k is the spring constant.
If the time period of oscillation of mass M is one second, we can write:
1 = 2π √(M/k)
Squaring both sides, we get:
1/4π² = M/k
Multiplying both sides by 4M, we get:
M = 4M/4π²
Now, we need to find the time period of 4M. Let's call this time period T'.
Using the formula for the time period, we get:
T' = 2π √(4M/k)
Substituting the value of M from the earlier equation, we get:
T' = 2π √[(4M)/(4π²)]
Simplifying, we get:
T' = 2π √(1/π²)
T' = 2 seconds
Therefore, the time period of 4M will be 2 seconds.

The correct answer is: 100

As it is given, T.E = K.E + P.E = 160 J

160 = K.Emax+ P.Emin (at mean position)

160 = 0+P.Emax

P.Emax = 160J

K.Emax = 1/2KA² = 1/2×2×106×(0.01)² = 100 J