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A body performing simple harmonic with amplitude of 10cm. The velocity of the body is triple by air jet when it is at 5cm from its mean position. The new Aplitude is $\sqrt{x} c m .$ The value of $x$ is

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answer is 700.

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Detailed Solution

$3 V = ω \sqrt{A^{2} - x^{2}}$

$V = ω \sqrt{10^{2} - 5^{2}} ..... 1$

$3 V = ω \sqrt{A^{2} - 5^{2}} ..... 2$

$3 ω \sqrt{10^{2} - 5^{2}} = ω \sqrt{A^{2} - 5^{2}}$

$A^{2} = 75 \times 9 + 25$

$9 (100 - 25) = A^{2} - 25$

$675 + 25 = 700$

$A = \sqrt{700}$

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