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  • A body is performing simple harmonic with an amplitude of . The velocity of the body was tripled by air Jet when it is at

A body is performing simple harmonic with an amplitude of 10 \mathrm{~cm}. The velocity of the body was tripled by air Jet when it is at 5 \mathrm{~cm} from its mean position. The new amplitude of vibration is \sqrt{x} \mathrm{~cm}. The value of  \mathrm{x} is____________.

Option: 1

700


Option: 2

200


Option: 3

300


Option: 4

250


Answers (1)

best_answer

\mathrm{\begin{aligned} &\mathrm{v=\omega \sqrt{A^{2}-x^{2}}} \\ &\text { Initially } \\ &\mathrm{v=w \sqrt{A^{2}-5^{2}} }\\ &\text { Given. } \text{A}=10 \mathrm{~cm} \\ &\text { As Given in question }\\ & \mathrm{3 \omega \sqrt{A^{2}-5^{2}}=\omega \sqrt{A'^{2}-5^{2}}} \\ &9 \times 75=\left(A^{'}\right)^{2}-25 \\ &A^{'}=\sqrt{28 \times 25}=\sqrt{700} \end{aligned}}

Hence the correct answer is 700

Posted by

Deependra Verma

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