Correct Answer is Option (A)

The point a covers 30${}^{\circ }$ in 0.1 sec.

Means $\frac{\pi }{6}\stackrel{}{⟶}0.1$ sec.

$1\stackrel{}{⟶}\frac{0.1}{\frac{\pi }{6}}$

$2\pi \stackrel{}{⟶}\frac{0.1\times 6}{\pi }\times 2\pi$

$T=1.2$ sec.

We know that $\omega =\frac{2\pi }{T}$

$\omega =\frac{2\pi }{1.2}$

Restoration force $(F)=m{\omega }^{2}A$

Then Restoration force per unit mass $\left(\frac{F}{m}\right)={\omega }^{2}A$

$\left(\frac{F}{m}\right)={\left(\frac{2\pi }{1.2}\right)}^2\times 0.36$

$\cong 9.87$ N

Correct Answer is Option (D)

Since, the particle is executing SHM.

Therefore, displacement equation of wave will be

$y=A\sin \omega t$

$\Rightarrow y/A=\sin \omega t$

and wave velocity equation will be

$v_y=\frac{dy}{dt}=A\omega \cos \omega t$

$\Rightarrow v_y/A\omega =\cos \omega t$

Now, ${\sin }^2\omega t+{\cos }^2\omega t=1$

$\therefore$ $(y/A{)}^2+(v_y/A\omega {)}^2=1$

This equation is similar to the equation of ellipse.

Correct Answer is (2)

x(t) = A sin($\omega$t + $\varphi$)

v(t) = A$\omega$ cos ($\omega$t + $\varphi$)

2 = A sin$\varphi$ ...... (1)

2$\omega$ = A$\omega$ cos$\varphi$ ....... (2)

From (1) and (2)

tan$\varphi$ = 1

$\varphi$ = 45${}^{\circ }$

Putting value of $\varphi$ in equation (1),

$2=A\left\{\frac{1}{\sqrt{2}}\right\}$

$A=2\sqrt{2}$

$\therefore$ x = 2

Correct Answer is (6)

Time period (T) = 2 sec.

X = A sin ($\omega$t + $\varphi$) ($\varphi$ = 0 at M.P.)

$\Rightarrow$ $\frac{A}{2}=A\sin \frac{2\pi }{T}t$

$\Rightarrow$ $\frac{2\pi }{2}t=\frac{\pi }{6}$

$\Rightarrow$ $t=\frac{1}{6}$

$\therefore$ a = 6

Correct Answer is (3)

For a particle executes S.H.M.

$V=\omega \sqrt{a^2-x^2}$

Given, $V=\frac{V_{max}}{2}=\frac{A\omega }{2}$

$\frac{A^2{\omega }^2}{4}={\omega }^2{a}^2-{\omega }^2{x}^2$

$x=\frac{\sqrt{3}}{2}a$