A particle performs SHM along x-axis with amplitude $A$ and period $T$. The time taken from $x = 0$ to $x = A/2$ is:
A$T/12$
B$T/8$
C$T/6$
D$T/4$
Answer & Solution
Correct answer: A. $T/12$
$x = A\sin(\omega t)$, starting from mean. At $x = A/2$: $\sin(\omega t) = 1/2$ ⇒ $\omega t = \pi/6$ ⇒ $t = \pi/(6\omega) = (\pi/6)/(2\pi/T) = T/12$.
Related questions
Two waves of slightly different frequencies but equal amplitude combine to produce:Two identical waves arrive crest on trough, so the amplitude of the result is:Two identical waves arrive crest on crest, so the amplitude of the result is:A wave whose disturbance is parallel to its direction of travel is:A wave whose disturbance is perpendicular to its direction of travel is:The distance between adjacent identical parts of a wave is its:Driving a system at its natural frequency, so the amplitude grows large, produces:A system that reaches equilibrium quickly but oscillates about it is described as: