JEE MAIN - Physics (2022 - 29th June Morning Shift - No. 21)
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 _____________.
Відповідь
700
Пояснення
$$v = \omega \sqrt {{A^2} - {y^2}} $$
$$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $$
$$ \Rightarrow 9 \times 75 = {(A')^2} - 25$$
$$ \Rightarrow A' = \sqrt {28 \times 25} $$ cm
$$ \Rightarrow x = 700$$