Question

A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $\sqrt x$ cm. The value of x is ___.

Answer 1

The correct option is: 700

$v=ω\sqrt{A^2-y^2}$

⇒ 9 × 75 = (A ′)2 – 25

$⇒A'=\sqrt{28×25}\,cm$

⇒ x = 700