Question

A particle performing S.H.M. has a K.E. of 480
erg when it is at 2 cm from mean position, and
420 ergs when it is at a distance of 4 cm from
mean position. Determine total energy and
amplitude of vibration.
​

Answer 1

Answer:

Total energy of oscillation is 500 erg and amplitude of vibration is 10 cm

Explanation:

As we know that kinetic energy of SHM is given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

here we have

KE = 480 erg at x = 2 cm

$480 = \frac{1}{2}m\omega^2(A^2 - 2^2)$

also we have

KE = 420 erg at x = 4 cm

$420 = \frac{1}{2}m\omega^2(A^2 - 4^2)$

now from above two equations we have

$\frac{480}{420} = \frac{(A^2 - 2^2)}{(A^2 - 4^2)}$

$8(A^2 - 16) = 7(A^2 - 4)$

$A^2 = 100$

$A = 10 cm$

Now we know that total energy of SHM is given as

$E = \frac{1}{2}m\omega^2 A^2$

$E = \frac{1}{2}m\omega^2(10^2)$

$E = 500 erg$

#Learn

Topic : Energy of vibration

https://brainly.in/question/11839706