SHMHard
Question
A particle performs S.H.M. of amplitude A with angular frequency ω along a straight line. When it is at a distance 
A from mean positive, its kinetic energy gets increased by an amout 
mω2A2 due to an impulsive force. What will be its new amplitude ?
A from mean positive, its kinetic energy gets increased by an amout mω2A2 due to an impulsive force. What will be its new amplitude ?Options
A.
A
AB.A
AC.√2 A
D.√5 A
Solution
at x = A
KE =
mω2 
mω2A2.
KE is increased by an amount of
mω2A2.Let now
amplitude be A1 then total Ke.
KE1 =
mω2A2 + mω2A2
=
mω2A2 
mω2
⇒ A1 = √2 A
AKE =
mω2 mω2A2.KE is increased by an amount of
mω2A2.Let nowamplitude be A1 then total Ke.
KE1 =
mω2A2 + mω2A2=
mω2A2 mω2 ⇒ A1 = √2 ACreate a free account to view solution
View Solution FreeMore SHM Questions
The distance between the points of suspension and centre of oscillation for a compound pendulum is :...A single of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultan...The frequency of a simple pendulum is n oscillations per minute while that of another is (n + 1) oscillations per minute...The position of a particle at time t moving in x-y plane is given by = (î + 2ĵ) A cos ωt. Then, the motio...A clock S works on the oscillations of a spring. Another clock P works on the oscillations of pendulum both the clocks k...