Correct Answer is Option (C)

$\begin{array}{rl}& \mathrm{x}=\mathrm{A}\sin (\omega \mathrm{t})\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}=\mathrm{A}\omega \cos (\omega \mathrm{t})\\ & \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{a}=-{\omega }^2\mathrm{A}\sin (\omega \mathrm{t})\\ & \mathrm{a}=-{\left(\frac{2\pi }{8}\right)}^2\times 1\sin (\frac{2\pi }{8}\times 2)\\ & \Rightarrow \mathrm{a}=-\frac{{\pi }^2}{16}\times \sin \left(\frac{\pi }{2}\right)\\ & \therefore \mathrm{a}=\frac{-{\pi }^2}{16}\mathrm{m}/{\mathrm{s}}^2\end{array}$

Correct Answer is Option (B)

Here,

$\sin \omega t$ and $\sin \left(\omega t+\frac{\pi }{4}\right)$ clearly suggest a periodic sinusoidal function (SHM)

And $\sin \omega t+\cos \omega t=\sqrt{2}\sin \left(\omega t+\frac{\pi }{4}\right)$ is also a periodic sinusoidal function (SHM)

While, $y=e^{-\omega t}$ is an exponentially decreasing function and thus does not have periodicity.

Correct Answer is Option (D)

Displacement, y = A.sin $\omega$ t
Velocity, $\upsilon$ = $\frac{dy}{dt}A\omega .\cos \omega t$
Acceleration, a = $-A{\omega }^2.\sin \omega t$
$=A{\omega }^2.\sin \left(\omega t+\pi \right)$
So the phase difference between y and a is $\pi$.

Correct Answer is Option (C)

Angular velocity, ($\omega$) = $\frac{2\pi }{T}=\frac{2\pi }{4}=\frac{\pi }{2}$ rad/s

Since, at t = 0, displacement (y) is maximum, so equation will be cosine function.

y = a cos $\omega$t

$\Rightarrow$ y = 3 cos $\left(\frac{\pi t}{2}\right)$

Correct Answer is Option (D)

Since, net displacement in complete cycle ∆y = 0

So, Average velocity

=

Correct Answer is Option (B)

Given, acceleration, a = 20 m/s²,
and displacement, y = 5m

Magnitude of acceleration of a particle
moving in a SHM is, |a| = ${\omega }^2$y; where y is amplitude.

$\Rightarrow$ 20 = ${\omega }^2$(5)

$\Rightarrow$ $\omega =2$ rad/s

Time period of pendulum,

T = $\frac{2\pi }{\omega }=\frac{2\pi }{2}$ = $\pi$ s

Correct Answer is Option (B)

Given, Amplitude A = 3 cm
When particle is at x = 2 cm
According to question, magnitude of velocity = acceleration

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

$\sqrt{{\left(3\right)}^2-{\left(2\right)}^2}=2\left(\frac{2\pi }{T}\right)$

$\sqrt{5}=\frac{4\pi }{T}\Rightarrow T=\frac{4\pi }{\sqrt{5}}$

Correct Answer is Option (B)

As, we know, in SHM
Maximum acceleration of the particle, $\alpha$ = A${\omega }^2$
Maximum velocity, $\beta$ = A$\omega$

$\Rightarrow \omega =\frac{\alpha }{\beta }$

$\Rightarrow T=\frac{2\pi }{\omega }=\frac{2\pi \beta }{\alpha }$   $[\because \omega =\frac{2\pi }{T}]$

Correct Answer is Option (D)

As we know, for particle undergoing SHM,

$V=\omega \sqrt{A^2-X^2}$

$V_1^2={\omega }^2\left(A^2-x_1^2\right)$

$V_2^2={\omega }^2\left(A^2-x_2^2\right)$

Substracting we get,

$\frac{V_1^2}{{\omega }^2}+x_1^2=\frac{V_2^2}{{\omega }^2}+x_2^2$

$\Rightarrow \frac{V_1^2-V_2^2}{{\omega }^2}=x_2^2-x_1^2$

$\Rightarrow w=\sqrt{\frac{V_1^2-V_2^2}{x_2^2-x_1^2}}$

$\Rightarrow T=2\pi \sqrt{\frac{x_2^2-x_1^2}{V_1^2-V_2^2}}$

Correct Answer is Option (C)

Here, X = Acos$\omega$t

$\therefore$ Velocity, $v=\frac{dX}{dt}=\frac{d}{dt}\left(A\cos \omega t\right)$

$=-A\omega \sin \omega t$

Acceleration, $a=\frac{dv}{dt}=\frac{d}{dt}\left(-A\omega \sin \omega t\right)$

$=-A{\omega }^2\cos \omega t$

Hence the variation of a with t is correctly shown by graph (c).

Correct Answer is Option (C)

As $\frac{v^2}{a^2{\omega }^2}+\frac{y^2}{a^2}=1$, This is the equation of ellipse. Hence the graph is an ellipse. P versus x graph is similar to V versus x graph.