Question

A simple harmonic motion of a spring-block system has an amplitude A & time period T. Consider x=0 for equilibrium position. The time required by the block to travel from $x=Atox=\frac{A}{2}$ is -

• A. $\frac{T}{6}$
• B. $\frac{T}{4}$
• C. $\frac{T}{3}$
• D. $\frac{T}{2}$

Answer 1

The correct option is A

$\frac{T}{6}$

Time required to travel from x = A to x = 0 is $\frac{T}{4}$ & time required to travel from x = 0 to x = $\frac{A}{2}$ is obtained from

$y=Asin\left(\frac{2\pi }{T}\right)t$

$\frac{A}{2}=Asin\left(\frac{2\pi }{T}\right)t$

$\Rightarrow sin\left(\frac{\pi }{6}\right)=sin\left(\frac{2\pi }{T}\right)t$

$\Rightarrow t=\frac{T}{12}$

$\Rightarrow$ Time required to travel from $x=Atox=\frac{A}{2}$ will be

$\frac{T}{4}-\frac{T}{12}=\frac{T}{6}$