⇒Oscillations

Correct Option is (B)

$\begin{array}{rl}& \mathrm{T}=\frac{2\pi }{\sqrt{3}}\mathrm{s},\mathrm{ZA}=4\mathrm{cm}\\ & \therefore \mathrm{A}=2\mathrm{cm}=2\times {10}^{-2}\mathrm{m}\\ & \text{ Acceleration, }a={\omega }^2\mathrm{x}\\ & \text{ Velocity, }v=\omega \sqrt{A^2-x^2}\\ & \therefore {\omega }^2\mathrm{x}=\omega \sqrt{A^2-{\mathrm{x}}^2}\\ & \omega x=\sqrt{A^2-x^2}\\ & {\omega }^2{x}^2=A^2-x^2\\ & \therefore 3{\mathrm{x}}^2={\mathrm{A}}^2-{\mathrm{x}}^2(\therefore \omega =\frac{2\pi }{\mathrm{T}}=\sqrt{3})\\ & \therefore 4{\mathrm{x}}^2={\mathrm{A}}^2\\ & \mathrm{c}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1\mathrm{cm}\\ & \end{array}$

Correct Option is (A)

At the highest point the particle will come to rest momentarily, hence it is at extreme position and has maximum force and acceleration. Since the spring is unstretched, the restoring forceto provided by the weight of the particle

$\therefore \mathrm{mA}{\omega }^2=\mathrm{mg}$

or $A{\omega }^2=\mathrm{g}$

$\begin{array}{rl}\therefore A& =\frac{g}{{\omega }^2}\\ \omega & =2\pi f=2\pi \times 5=10\pi \\ \therefore A& =\frac{10}{100{\pi }^2}=\frac{1}{10{\pi }^2}\\ V_{max}& =A\omega =\frac{1}{10{\pi }^2}\times 10\pi =\frac{1}{\pi }\end{array}$

Correct Option is (B)

$x=x_1+x_2=A\sin (\omega t+\alpha )$

Maximum value of $x_1+x_2=A=\sqrt{2}a$

$\begin{array}{rl}& A^2=a^2+a^2+2a^2\cos \alpha =2a^2(1+\cos \alpha )\\ & \therefore 2a^2=2a^2(1+\cos \alpha )\therefore 1=1+\cos \alpha \\ & \therefore \cos \alpha =0\therefore \alpha =\frac{\pi }{2}\end{array}$

Correct Option is (A)

m = 5 g = 5 $\times$ 10${}^{-3}$ kg, A = 0.3 m, T = $\frac{\pi }{5}$s

$\omega =\frac{2\pi }{\mathrm{T}}=10$ rad/s

Maximum force

F = m${\omega }^2$A = 5 $\times$ 10${}^{-3}\times (10{)}^2\times 0.3=0.15$ N