A body is in simple harmonic motion with time period half second $(T\, = 0.5\, s)$ and amplitude one $cm\, (A\,= 1\, cm)$. Find the average velocity in the interval in which it moves form equilibrium position to half of its amplitude .... $cm/s$
JEE MAIN 2014, Medium
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Given: Time period, $T=0.5$ sec
Amplitude, $\mathrm{A}=1 \mathrm{cm}$
Average velocity in the interval in which body moves from equilibrium to half of its amplitude, $\mathrm{v}=?$
Time taken to a displacement $A/2$ where $A$ is the amplitude of oscillation from the
mean position $^{\prime} \mathrm{O}^{\prime}$ is $\frac{\mathrm{T}}{12}$
Therefore, time, $t=\frac{0.5}{12} \mathrm{sec}$
Displacement, $s=\frac{A}{2}=\frac{1}{2} c m$
Average velocity, $v=\frac{\frac{A}{2}}{t}=\frac{\frac{1}{2}}{\frac{0.5}{12}}=12 \mathrm{cm} / \mathrm{s}$
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