
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
$y = \frac{1}{{\sqrt a }}\,\sin \,\omega t \pm \frac{1}{{\sqrt b }}\,\cos \,\omega t$ will be
$(a)$ Potential energy is always equal to its $K.E.$
$(b)$ Average potential and kinetic energy over any given time interval are always equal.
$(c)$ Sum of the kinetic and potential energy at any point of time is constant.
$(d)$ Average $K.E.$ in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below:

