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 (integer)
700
Solution
<p>$v = \omega \sqrt {{A^2} - {y^2}}$</p>
<p>$$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $$</p>
<p>$\Rightarrow 9 \times 75 = {(A')^2} - 25$</p>
<p>$\Rightarrow A' = \sqrt {28 \times 25}$ cm</p>
<p>$\Rightarrow x = 700$</p>
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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