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13.Oscillations
medium
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 ..... $\sec$ ?
A
$1$
B
$2$
C
$3$
D
$4$
Solution
$X=A \sin \omega t$
or $10=20 \sin \omega t$
$\therefore \omega t=\left(\frac{\pi}{6}\right) \Rightarrow t=\frac{\pi}{6 \omega}$
The desired time will be $2 t$
or $\frac{\pi}{3 \omega}=\frac{\pi}{3(2 \pi / T)}=\frac{T}{6}=\frac{12}{6}=2 s$
Standard 11
Physics
