Correct Answer is (40)

$F=-25x$

$.250\frac{d^{2}x}{d{t}^2}=-25x$

$\frac{d^{2}x}{d{t}^2}=-100x$

$\Rightarrow \omega =10$ rad/sec

& $\omega A=v_{max}$

$10A=4$

$\Rightarrow A=0.4$ m

$=40$ cm

Correct Answer is Option (B)

Let $x=A\sin \omega t$

$\Rightarrow v=A\omega \cos \omega t$

$\Rightarrow v=\pm \omega \sqrt{A^2-x^2}$

$\Rightarrow \frac{v^2}{{\omega }^2}+x^2=A^2$

$\Rightarrow$ Ellipse

Correct Answer is Option (A)

$x=4\sin \left(\frac{\pi }{2}-\omega t\right)$

$=4\cos (\omega t)$

$y=4\sin (\omega t)$

$\Rightarrow x^2+y^2=4^2$

$\Rightarrow$ The particle is moving in a circular motion with radius of 4 m.

Correct Answer is Option (B)

$x=\sin \left(\pi t+\frac{\pi }{3}\right)m$

$\Rightarrow \frac{dx}{dt}=\pi \cos \left(\pi t+\frac{\pi }{3}\right)$

$=\pi \cos \left(\pi +\frac{\pi }{3}\right)$ at $t=1s$

$=-\frac{\pi }{2}$ m/s

or $\left|\frac{dx}{dt}\right|=157$ cm/s

Correct Answer is Option (D)

Time taken by the harmonic oscillator to move from mean position to half of amplitude is $\frac{T}{12}$

So, $\frac{T}{12}$ = 3

T = 36 sec.