Question 79: Oscillations

Question

T is the time period of simple pendulum on the earth's surface. Its time period becomes $$x$$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $$x$$ will be :

Accepted Answer

At surface of earth time period
$$
\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}
$$
At height $\mathrm{h}=\mathrm{R}$
$$
\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}ight)^2}=\frac{\mathrm{g}}{4} \\
& 	herefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\
& \Rightarrow \mathrm{xT}=2 	imes 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\
& \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2
\end{aligned}
$$

Answer

4

Answer

$$\frac{1}{2}$$

Answer

2

Answer

$$\frac{1}{4}$$

Oscillations

T is the time period of simple pendulum on the earth's surface. Its time period becomes $x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x$ will be :

Choose the correct answer: