
direction and elongation in spring $=\mathrm{x}_{1}$ then
$x \cos \theta=x_{1}$ $...(1)$
Restoring force $\mathrm{F}=2 \mathrm{kx}_{1} \cos \theta$
$\mathrm{F}=2 \mathrm{k} \cos ^{2} \theta \times$
Hence $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \cos ^{2} \theta}}=2 \pi \sec \theta \sqrt{\frac{\mathrm{m}}{2 \mathrm{k}}}$
$y = 0.2\sin \left( {10\pi t + 1.5\pi } \right)\cos \left( {10\pi t + 1.5\pi } \right)$
The motion of particle is
