Let the $$SHM’s$$  be
$$\eqalign{ & x = a\sin \omega t\,......\left( {\text{i}} \right) \cr & {\text{and}}\,\,y = b\sin \left( {\omega t + \pi } \right) \cr & {\text{or}}\,\,y = - b\sin \omega t\,......\left( {{\text{ii}}} \right) \cr} $$
From Eqs. (i) and (ii)
$$\eqalign{ & \frac{x}{a} = \sin \omega t\,\,{\text{and}}\,\, - \frac{y}{b} = \sin \omega t \cr & \therefore \frac{x}{a} = - \frac{y}{b}\,\,{\text{or}}\,\,y = - \frac{b}{a}x \cr} $$
It is an equation of a straight line.

The net force becomes zero at the mean point. Therefore, linear momentum must be conserved.
$$\eqalign{ & \therefore M{v_1} = \left( {M + m} \right){v_2} \cr & M{A_1}\sqrt {\frac{k}{M} = } \left( {M + m} \right){A_2}\sqrt {\frac{k}{{m + M}}} \,\therefore \left( {V = A\sqrt {\frac{k}{M}} } \right) \cr & {A_1}\sqrt M = {A_2}\sqrt {M + m} \,\therefore \,\frac{{{A_1}}}{{{A_2}}} = \sqrt {\frac{{m + M}}{M}} \cr} $$

In simple harmonic motion, the displacement equation is, $$x = a\sin \omega t$$
where, $$a$$ is the amplitude of the motion.
$$\eqalign{ & {\text{Velocity, }}v = \frac{{dx}}{{dt}} = a\omega \cos \omega t \cr & v = a\omega \sqrt {1 - {{\sin }^2}\omega t} \cr & v = \omega \sqrt {{a^2} - {x^2}} \,......\left( {\text{i}} \right) \cr & {\text{Acceleration, }}\alpha = \frac{{dv}}{{dt}} = \frac{d}{{dt}}\left( {a\omega \cos \omega t} \right) \cr & \alpha = - a{\omega ^2}\sin \omega t \cr & \alpha = - {\omega ^2}x\,......\left( {{\text{ii}}} \right) \cr & {\text{When}}\,x = 0,v = a\omega = {v_{\max }} \cr & \alpha = 0 = {\alpha _{\min }} \cr & {\text{When}}\,x = a,v = 0 = {v_{\min }} \cr & \alpha = - {\omega ^2}a = {\alpha _{\max }} \cr} $$
Hence, it is clear that when $$v$$ is maximum, then $$\alpha $$ is minimum (i.e. zero) or vice-versa.

$${t_1} = 2\pi \sqrt {\frac{m}{{{k_1}}}} ,{t_2} = 2\pi \sqrt {\frac{m}{{{k_2}}}} $$
when springs are in series then,
$$\eqalign{ & {k_{{\text{eff}}}} = \frac{{{k_1}{k_2}}}{{{k_1} + {k_2}}}\,\,\therefore T = 2\pi \sqrt {\frac{{m\left( {{k_1} + {k_2}} \right)}}{{{k_1}{k_2}}}} \cr & \therefore T = 2\pi \sqrt {\frac{m}{{{k_2}}} + \frac{m}{{{k_1}}}} = 2\pi \sqrt {\frac{{t_2^2}}{{{{\left( {2\pi } \right)}^2}}} + \frac{{t_1^2}}{{{{\left( {2\pi } \right)}^2}}}} \cr & \Rightarrow {T^2} = t_1^2 + t_2^2 \cr} $$

Velocity of the particle executing $$SHM$$  at any instant is defined as the time rate of change of its displacement at that instant. It is given by
$$v = \omega \sqrt {\left( {{a^2} - {x^2}} \right)} $$
where, $$x$$ is displacement of the particle is acceleration and $$\omega $$ is angular frequency.
Case I
$$10 = \omega \sqrt {{a^2} - 16} \,......\left( {\text{i}} \right)$$
Case II
$$8 = \omega \sqrt {{a^2} - 25} \,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i), we get
$$\eqalign{ & \frac{5}{4} = \frac{{\sqrt {{a^2} - 16} }}{{\sqrt {{a^2} - 25} }}\,\,{\text{or}}\,\,\frac{{25}}{{16}} = \frac{{{a^2} - 16}}{{{a^2} - 25}} \cr & {\text{or}}\,\,16\,{a^2} - 256 = 25\,{a^2} - 625 \cr & {\text{or}}\,\,{a^2} = \frac{{369}}{9} \cr} $$
Putting value of $${a^2}$$ in Eg. (i), we get
$$\eqalign{ & 10 = \omega \sqrt {\left( {\frac{{369}}{9} - 16} \right)} \cr & {\text{or}}\,\,\omega = \frac{{10 \times 3}}{{15}} = 2\,rad/s \cr} $$
$$\therefore $$ Time period $$T = \frac{{2\pi }}{\omega } = \frac{{2\pi }}{2} = \pi \sec $$

The second pendulum placed in a space laboratory orbiting around the earth is in a weightlessness state. Hence $$g = 0$$  so $$T = \infty $$

$${\text{Given}}\,{\text{that,}}\,\,\frac{{{A_{\max }}}}{{{v_{\max }}}} = 10\,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,\omega = 10\,{s^{ - 1}}$$
Displacement is given by
$$\eqalign{ & x = a\sin \left( {\omega t + \frac{\pi }{4}} \right) \cr & {\text{at}}\,\,t = 0,x = 5;5 = a\sin {45^ \circ } \Rightarrow a = 5\sqrt 2 \cr & {A_{\max }} = a{\omega ^2} = 500\sqrt 2 \,m/{s^2} \cr} $$

$${\text{Potential energy}} \propto \left( {{\text{Amplitude}}} \right) \propto {\left( {{\text{Displacement}}} \right)^{\text{2}}}{\text{ }}$$
$$P.E.$$  is maximum at maximum distance.
$$P.E.$$  is zero at equilibrium point.
$$P.E.$$  curve is parabolic.

Apply vector formula to determine resultant acceleration of the both.

The bob is now under the combined action of two accelerations, $$g$$ vertically downwards and $$\alpha $$ along the horizontal.
∴ Resultant acceleration
$$g' = \sqrt {{g^2} + {\alpha ^2}} $$

If $$l$$ is the length of pendulum and $$\theta $$ the angular amplitude, then height

$$\eqalign{ & h = AB - AC \cr & = l - l\cos \theta \cr & = l\left( {1 - \cos \theta } \right)\,......\left( {\text{i}} \right) \cr} $$
At point $$P$$ (maximum displacement position i.e. extreme position), potential energy is maximum and kinetic energy is zero. At point $$B$$ (mean or equilibrium position) potential energy is minimum and kinetic energy is maximum, so from principle of conservation of energy.
$$\eqalign{ & \left( {PE + KE} \right)\,{\text{at}}\,P = \left( {KE + PE} \right)\,{\text{at}}\,B \cr & {\text{or}}\,\,mgh + 0 = \frac{1}{2}m{v^2} + 0 \cr & {\text{or}}\,\,v = \sqrt {2gh} \,......\left( {{\text{ii}}} \right) \cr} $$
Substituting the value of $$h$$ from Eq. (i) into Eq. (ii),
$$v = \sqrt {2gl\left( {1 - \cos \theta } \right)} $$