Question

The time period of a particle executing SHM is $8$s. At $t=0$, it is at mean position. If A is the amplitude of particle, then the distance travelled by particle (in second) is?

• A. $\sqrt{2}A$
• B. $\left(1-\frac{1}{\sqrt{2}}\right)A$
• C. $2A$
• D. $\frac{A}{\sqrt{2}}$

Answer 1

Answer: (B)

$\left(1-\frac{1}{\sqrt{2}}\right)A$

The displacement of a particle executing SHM with initial phase zero is given by
$y=A\sin \omega t=A\sin \left(\frac{2\pi }{T}t\right)$
when, $t=1$s
$y_1=A\sin \left(\frac{2\pi }{8}\times 1\right)$
$=A\times \sin \left(\frac{\pi }{4}\right)=\frac{A}{\sqrt{2}}$
when, $t=2$s
$y_2=A\sin \left(\frac{2\pi }{8}\times 2\right)$
$=A\sin \frac{\pi }{2}=A$
So, the distance travelled in $2$nd second is
$y=y_2-y_1$
$=A-\frac{A}{\sqrt{2}}=(1-\frac{1}{\sqrt{2}})A$.