Find the time period of oscillations of a body placed in a tunnel dug into earth anywhere in the earth (approximately)             

If ‘y’ is the distance of the body from the centre of the tunnel then

$y=x\sin \theta \cdots \left(1\right)$

The restoring force

$\text{F}=-\left({\text{F}}_{\text{g}}\sin \theta \right)$

$\Rightarrow \text{F}=-\frac{{\text{GM}}^{\text{1}}\text{m}}{x^2}\sin \theta$

$\text{F}=-\text{G}\left(\frac{\text{M}}{\frac{4}{3}\pi {\text{R}}^{\text{3}}}\right)\frac{\left(\frac{4}{3}\pi x^3\right)\text{m}\sin \theta }{x^{\text{2}}}$

$\text{F}=-\frac{\text{GMm}x\sin \theta }{{\text{R}}^{\text{3}}}$

$\text{a}=\frac{\text{F}}{\text{m}}$

$\text{a}=\frac{\text{GM}}{{\text{R}}^{\text{3}}}x\sin \theta$

Substituting$\frac{\text{GM}}{{\text{R}}^{\text{2}}}=\text{g}\text{and}x\sin \theta =y$  

We get$\text{a}=\frac{\text{g}}{\text{R}}\left(-y\right)=-\left(\frac{\text{g}}{\text{R}}\right)y;\text{a}=-{\omega }^{2}y$

We get$\omega =\sqrt{\frac{\text{g}}{\text{R}}}\Rightarrow \text{T}=2\pi \sqrt{\frac{\text{R}}{\text{g}}}$

$\text{T}=2\pi \sqrt{\frac{6400\times 1000}{10}}$

$\Rightarrow \text{T}=1600\pi$