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  • The velocity of a particle in S.H.M at position Y1 and Y2 are v1 and v2 respectively. Determine the value of time period and amplitude. &nbsp

The velocity of a particle in S.H.M at position Y1 and Y2 are v1 and v2 respectively. Determine the value of time period and amplitude.
 

Option: 1

\sqrt{\frac{y_2^2-y_1^2}{v_1^2-v_2^2}}


Option: 2

2 \pi \sqrt{\frac{y_2^2-y_1^2}{v_1^2-v_2^2}}


Option: 3

\pi \sqrt{\frac{y_2^2-y_1^2}{v_1^2-v_2^2}}


Option: 4

2 \pi \left ({\frac{y_2^2-y_1^2}{v_1^2-v_2^2}} \right )


Answers (1)

best_answer

Y1 → v1

Y2 → v2

\begin{aligned} & v=w \sqrt{A^2-x^2} \\ & v_1=w \sqrt{A^2-y_1^2} \\ & v_2=w \sqrt{A^2-y_2^2}\\ \end{aligned}\\ \text{(1) Divide by (2)}\\\\ \begin{aligned} & \frac{v_1}{v_2}=\frac{w \sqrt{A^2-y_1 2}}{w \sqrt{A^2-y_2{ }^2}} \\ & \frac{v_1^2}{v_2^2}=\frac{A^2-y_1^2}{A^2-y_2^2} \\ & v_1^2 A^2-v_1^2 y_2^2=v_2^2 A^2-v_2^2 y_1^2 \\ & v_1^2 A^2-v_2^2 A^2=v_1^2 y_2^2-v_2^2 y_1^2 \\ & \left(v_1^2-v_2^2\right) A^2=v_1^2 y_2^2-v_2^2 y_1^2 \\ & A^2=\frac{v_1^2 y_2^2-v_2^2 y_1^2}{v_1^2-v_2^2}\\ \end{aligned}\\ A=\sqrt{\frac{v^2 y_2^2-v_2^2 y_1^2}{v^2-v_2^2}}\\\\ \text{for time Period}\\\\ \begin{aligned} & v_1^2=w^2\left(A^2-y_1^2\right) \\ & \frac{v_1^2}{w^2}=A^2-y_1^2 \\ & \frac{v_1^2}{A^2-y_1^2}=w^2 \\ & w=\sqrt{\frac{V_1^2}{A^2-x_1^2}} \\ & w=\sqrt{\frac{v_1^2}{\frac{v_1^2 y_2{ }^2-v_2^2 y_1{ }^2}{v_1{ }^2-v_2{ }^2}}-y_1^2} \\ & =\sqrt{\frac{v_1{ }^2}{\frac{v_1^2 y_2{ }^2-v_2^2 y_1^2-v_1^2 y_1{ }^2+v_2{ }^2 y_1{ }^2}{v_1{ }^2-v_2 2}}} \\ & \end{aligned}\\

\begin{aligned} & =\sqrt{\frac{v_1{ }^2\left(v_1{ }^2-v_2{ }^2\right)}{v_1{ }^2\left(y_2{ }^2-y_1{ }^2\right)}} \\ & \frac{2 \pi}{T}=\sqrt{\frac{v_1{ }^2-v_2 2}{y_2{ }^2-y_1{ }^2}} \\ & T=2 \pi \sqrt{\frac{y_2{ }^2-y_1{ }^2}{v_1{ }^2-v_2 2}} \end{aligned}

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vinayak

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