If displacement of particle is $$y,$$ then
$$\eqalign{ & KE = \frac{1}{2}m{\omega ^2}\left( {{a^2} - {y^2}} \right)\,\,\& \,P.E. = \frac{1}{2}m{\omega ^2}{y^2} \cr & {\text{If}}\,\,KE = PE\,\,{\text{then}}\,\,\frac{1}{2}m{\omega ^2}{y^2} \cr & = \frac{1}{2}m{\omega ^2}{a^2} - \frac{1}{2}m{\omega ^2}{y^2} \cr & \Rightarrow 2{y^2} = {a^2}\,\,\therefore y = \frac{a}{{\sqrt 2 }} \cr} $$

Centre of mass of combination of liquid and hollow portion (at position $$\ell $$), first goes down $$\left( {{\text{to}}\,\ell + \Delta \ell } \right)$$   and when total water is drained out, centre of mass regain its original position (to $$\ell $$),
$$T = 2\pi \sqrt {\frac{\ell }{g}} $$
$$\therefore 'T'$$ first increases and then decreases to original value.

$$\eqalign{ & T = 2\pi \sqrt {\frac{\ell }{g}} \,{\text{and}}\,T' = 2\pi \sqrt {\frac{{1.21\ell }}{g}} \,\left( {\because \ell ' = \ell + 21\% \,{\text{of}}\,\ell } \right) \cr & \% \,{\text{increase}}\, = \frac{{T' - T}}{T} \times 100 = \frac{{\sqrt {1.21\ell } - \sqrt \ell }}{{\sqrt \ell }} \times 100 = 10\% \cr} $$

As shown in the figure, $$g\sin \alpha $$  is the pseudo acceleration applied by the observer in the accelerated frame

$$\eqalign{ & {a_y} = g - g{\sin ^2}\alpha = g\left( {1 - {{\sin }^2}\alpha } \right) = g{\cos ^2}\alpha \cr & a = \sqrt {a_x^2 + a_y^2} \cr & = \sqrt {{g^2}{{\sin }^2}\alpha {{\cos }^2}\alpha + {g^2}{{\cos }^4}\alpha } \cr & = g\cos \alpha \sqrt {{{\sin }^2}\alpha + {{\cos }^2}\alpha } = g\cos \alpha \cr & \therefore T = 2\pi \sqrt {\frac{L}{{g\cos \alpha }}} \cr} $$
NOTE: Whenever point of suspension is accelerating use $$T = 2\pi \sqrt {\frac{L}{{{g_{eff{\text{ }}}}}}} $$

Oscillations along spring length are independent of gravitation.

In harmonic oscillator, the energy is maximum at $${\omega _2} = {\omega _0}$$  and amplitude is maximum at frequency $${\omega _1} < {\omega _0}$$  in the presence of damping, so $${\omega _1} \ne {\omega _0}$$  and $${\omega _2} = {\omega _0}.$$

$$\eqalign{ & K.E. = \frac{1}{2}m{a^2}{\omega ^2}{\cos ^2}\omega t \cr & {\text{and}}\,\,T.E. = \frac{1}{2}m{a^2}{\omega ^2} \cr & {\text{Given}}\,\,K.E. = 0.75\,T.E. \cr & \Rightarrow 0.75 = {\cos ^2}\omega t \Rightarrow \omega t = \frac{\pi }{6} \cr & \Rightarrow t = \frac{\pi }{{6 \times \omega }} \Rightarrow t = \frac{{\pi \times 2}}{{6 \times 2\pi }} \Rightarrow t = \frac{1}{6}s \cr} $$

$${p_{\max }} = \sqrt {2m{E_{\max }}} $$

As we know that, the condition for a body executing $$SHM$$  is $$F = - kx$$
$$\eqalign{ & {\text{So,}}\,a = \frac{F}{m} = - \frac{k}{m}x \cr & {\text{or}}\,a = - {\omega ^2}x \cr & {\text{Acceleration}} \propto - \left( {{\text{displacement}}} \right) \cr & A \propto - y \cr & A = - {\omega ^2}y \cr & A = - \frac{k}{m}y \cr & A = - ky \cr & {\text{Here,}}\,y = x + a \cr & \therefore {\text{Acceleration}} = - k\left( {x + a} \right) \cr} $$

In forced oscillations, the body oscillates at the angular frequency of the driving force.