JEE Mains · Physics · STD 11 - 13. 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________
- A \(900\)
- B \(800\)
- C \(100\)
- D \(700\)
Answer & Solution
Correct Answer
(D) \(700\)
Step-by-step Solution
Detailed explanation
\(A =10 \,cm\) \(\therefore\) Total Energy \(=\frac{1}{2} KA ^{2}\) By energy conservation we can final \(v\) at \(x =5\) \(\frac{1}{2} K (10)^{2}=\frac{1}{2} K (5)^{2}+\frac{1}{2} mv ^{2}\) \(V =\sqrt{\frac{75 K }{ m }}\) Now, velocity is tripled through external mean so the…
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