Question

A particle is performing shm with an energy of vibration 90j and amplitude 6cm. When the particle reaches at distance 4cm from mean position, it is stopped for a moment and then released. The new energy of vibration will be:

Answer 1

we know, total energy of particle performing simple harmonic motion is directly proportional to square of amplitude.

so, $\frac{E_1}{E_2}=\frac{A_1^2}{A_2^2}$

here, $E_1$ = 90J , $A_1=6cm$ and $A_2=4cm$

now, 90/$E_2$ = 6²/4² = 9/4

or, $E_2$ = 4/9 × 90 = 40J

hence, the new energy of vibrations will be 40J

Answer 2

Answer:

40j

Explanation:

See the attachment