Physics - SHM Question with Solution | TestHub

The initial time period of the simple pendulum is given by:

$T_1=2\pi \sqrt{\frac{L}{g}}$

The point of suspension is moved upward according to the equation $y=k{t}^2$. The acceleration of the point of suspension is:

$a=\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}(2kt)=2k$

Given $k=1{\text{ m/s}}^2$, the upward acceleration is:

$a=2(1)=2{\text{ m/s}}^2$

When the point of suspension accelerates upward, the effective acceleration due to gravity becomes:

$g_{eff}=g+a=10+2=12{\text{ m/s}}^2$

The new time period $T_2$ is:

$T_2=2\pi \sqrt{\frac{L}{g_{eff}}}$

Squaring both time periods and taking their ratio:

$\frac{T_1^2}{T_2^2}=\frac{\frac{4{\pi }^{2}L}{g}}{\frac{4{\pi }^{2}L}{g_{eff}}}=\frac{g_{eff}}{g}$

Substituting the values of $g_{eff}$ and $g$:

$\frac{T_1^2}{T_2^2}=\frac{12}{10}=\frac{6}{5}$

Answer: $\frac{6}{5}$