Physics - SHM Question with Solution | TestHub

The equation of a particle performing linear S.H.M. starting from the mean position is given by:

$x=A\sin (\omega t)$

Given the period $T=8$ s, the angular frequency is:

$\omega =\frac{2\pi }{T}=\frac{2\pi }{8}=\frac{\pi }{4}$ rad/s

The distance travelled by the particle in the $1^{\text{st}}$ second (from $t=0$ to $t=1$ s) is:

$d_1=x(1)-x(0)=A\sin \left(\frac{\pi }{4}\times 1\right)-0=\frac{A}{\sqrt{2}}$

The distance travelled by the particle in the $2^{\text{nd}}$ second (from $t=1$ to $t=2$ s) is:

$d_2=x(2)-x(1)=A\sin \left(\frac{\pi }{4}\times 2\right)-A\sin \left(\frac{\pi }{4}\times 1\right)$

$d_2=A\sin \left(\frac{\pi }{2}\right)-\frac{A}{\sqrt{2}}=A(1)-\frac{A}{\sqrt{2}}=A\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$

The ratio of the distances travelled in the $1^{\text{st}}$ and $2^{\text{nd}}$ second is:

$\frac{d_1}{d_2}=\frac{\frac{A}{\sqrt{2}}}{A\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}=\frac{1}{\sqrt{2}-1}$

Thus, the ratio is $1:(\sqrt{2}-1)$.

Answer: $1:(\sqrt{2}-1)$