Home
Class 11
PHYSICS
The total mechanical energy of a spring ...

The total mechanical energy of a spring mass system in simple harmonic motion is `E=1/2momega^2 A^2`. Suppose the oscillating particle is replaced by another particle of double the mass while the amplitude A remains the same. The new mechanical energy will

A

become `2E`

B

becoem `E/2`

C

becoem `sqrt2E`

D

remain E

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

A particle at the end of a spring executes simple harmonic motion with a period t_(1) while the corresponding period for another spring is t_(2) if the oscillation with the two springs in series is T then

A particle executes simple harmonic motion with a period of 16s . At time t=2s , the particle crosses the mean position while at t=4s , its velocity is 4ms^-1 amplitude of motion in metre is

If particle is excuting simple harmonic motion with time period T, then the time period of its total mechanical energy is :-

Consider two satellites A and B of equal mass m, moving in the same circular orbit of radius r around the earth E but in opposite sense of rotation and therefore on a collision course (see figure) (i). In terms of G,M_(e) m and r find the total mechanical energy E_(A)+E_(B) of the two satellite plus earth system before collision (ii). If the collision is completely inelastic so that wreckage remains as one piece of tangled material (mass=2m) find the total mechanical energy immediately after collision. (iii). Describe the subsequent motion of the wreckage.

The motion of a particle of mass m is described by y = ut + 1/2 "gt"^2 . Find the force acting on the particle.

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is……….(where E is total (mechanical) energy)

A particle of mass m executes simple harmonic motion with amplitude a and frequency v. The average kinetic energy during its motion from the position of equilinrium to the end is.

A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance (2A)/3 from equilibrium position. The new amplitude of the motion is:

A particle executes simple harmonic motion according to equation 4(d^(2)x)/(dt^(2))+320x=0 . Its time period of oscillation is :-

A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is…….