Question 184: Oscillations

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 _____________.

Accepted Answer

$$v = \omega \sqrt {{A^2} - {y^2}} $$
$$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}}  = \omega \sqrt {{{(A')}^2} - {5^2}} $$
$$ \Rightarrow 9 	imes 75 = {(A')^2} - 25$$
$$ \Rightarrow A' = \sqrt {28 	imes 25} $$ cm
$$ \Rightarrow x = 700$$

Oscillations

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 _____________.

Integer / Numeric Answer Type