Correct Answer is Option (C)

The correct options are:

(A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM)

(B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction)

(C) The velocity is maximum at mean position. - True (the velocity is zero at the extreme positions and reaches a maximum at the mean position)

(D) The acceleration is minimum at extreme points. - False (the acceleration is maximum at the extreme points and zero at the mean position)

Correct Answer is Option (D)

The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$.

If we substitute these initial conditions into the equation for the displacement of the particle:

$x=A\sin (wt+\delta )$

We have:

$\frac{A}{2}=A\sin (\delta )$

Dividing both sides by $A$ gives us:

$\frac{1}{2}=\sin (\delta )$

The angle whose sine is $\frac{1}{2}$ is $\delta =\frac{\pi }{6}$ radians (or 30 degrees).

Therefore, the correct answer is $\delta =\frac{\pi }{6}$.

Correct Answer is Option (C)

After time $t$, the angular displacement will be

$\theta =\omega t$

Total angular displacement from $x$-axis.

${\theta }_{\text{total }}=\omega t+\frac{\pi }{6}(\because {30}^{\circ }=\frac{\pi }{6})$

Now, OP has two component

The horizontal component will be the projection along $x$-axis

$=r\cos \left({\theta }_{\text{Total }}\right)=r\cos (\omega t+\frac{\pi }{6})$

Correct Answer is Option (A)

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

$\begin{array}{rl}& x=\frac{A}{2}=A\sin (\omega t)\\ \\ & \frac{1}{2}=\sin (\omega t)\\ \\ & t=\left(\frac{\pi }{6\omega }\right)=2\\ \\ & \frac{\pi }{\omega }=12\sec \end{array}$

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

$\omega t=\left(\frac{\pi }{2}\right)$

$t=\left(\frac{\pi }{2\omega }\right)=6$ second

time $=6-2=4$ seconds

Correct Answer is (88)

$4v^2=50-x^2$

or $v=\frac{1}{2}\sqrt{50-x^2}$

Comparing the above equation with $v=\omega \sqrt{A^2-x^2}$

$\Rightarrow \omega =\frac{1}{2}$

& $A=\sqrt{50}$

so $\frac{2\pi }{T}=\frac{1}{2}$

$\Rightarrow T=4\pi \sec$

$=4\times \frac{22}{7}\sec$

$T=\frac{88}{7}\sec$

so $x=88$

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.