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Q. A particle executing SHM of amplitude $4 \, cm$ and $T=4 \, s$ . The time taken by it to move from positive extreme position to half the amplitude is

NTA AbhyasNTA Abhyas 2022

Solution:

Equation of motion $y=acos ωt$
$\Rightarrow \, \frac{a}{2}=acos ωt$
$\Rightarrow cos ωt=\frac{1}{3}$
$\Rightarrow \, ωt=\frac{\pi }{3}$
$\Rightarrow \frac{2 \pi t}{T}=\frac{\pi }{3}$
$\Rightarrow \, t=\frac{\frac{\pi }{3} \times t}{2 \pi }=\frac{4}{3 \times 2}=\frac{2}{3} \, s$