A particle executes SHM with amplitude of 20 | vedclass

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Question

$X=A \sin \omega t$

or $10=20 \sin \omega t$

$	herefore \omega t=\left(\frac{\pi}{6}ight) \Rightarrow t=\frac{\pi}{6 \omega}$

The desired

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$X=A \sin \omega t$

or $10=20 \sin \omega t$

$	herefore \omega t=\left(\frac{\pi}{6}ight) \Rightarrow t=\frac{\pi}{6 \omega}$

The desired time will be $2 t$

or $\frac{\pi}{3 \omega}=\frac{\pi}{3(2 \pi / T)}=\frac{T}{6}=\frac{12}{6}=2 s$

A particle executes $SHM$ with amplitude of $20 \,cm$ and time period is $12\, sec$. What is the minimum time required for it to move between two points $10\, cm$ on either side of the mean position ..... $\sec$ ?

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