Answer: Option B. -> False
:
B
So let's try to find equilibrium. But clearly, there is none.
At every point the earth is acted upon by Sun's gravitational pull.

Answer: Option C. -> 2f
:
C
During one complete oscillation, the kinetic energy will become maximum twice. Therefore the frequency of kinetic energy will be 2f.

Answer: Option B. -> 100 kg/s2
:
B
A general sinusoidal, oscillation of amplitude A, angular frequency $\omega andataphase\psi$ is given as
$x\left(t\right)=Asin(\omega t+\psi )$,
Compare this with the equation under discussion
$x=\left[\right(0.2cm\left)sin\right(100s^{-1})t+\frac{\pi }{6}]$
It is clear from a comparison, that
$A=0.2cm=0.002m$
$\omega =100s^{-1},and\psi =\frac{\pi }{6}$
We know that for oscillations in a spring mass system, the spring constant is related to the mass m and ω as
$k=m{\omega }^2$
The mass is given to us, viz. $m=10g=0.01kg$
$\therefore Thek=[0.01\times 1002]kg{s}^{-2}$
$=100kg{s}^{-2}$
Hence correct option is (b).

Answer: Option D. -> Zero
:
D
In one time period the particle comes back to its starting position, irrespective of from where did it start. So the displacement has to be 0.

Answer: Option B. -> yes, at any time of the day
:
B
Well! When the sun is exactly above the vertical circle, the shadow is formed and it does SHM very much as we know from the example
in the video linking SHM to circular motion. Question here actually is "will the shadow's motion be a SHM when the sun is at some angle (say $\theta$) ?”
Let's explore that

Let's say that the sun light from the sun makes an angle θ with the horizontal as you can see is the diagram. Since the creature moves with constant speed, the time it takes to move from A to B is the same it takes to move from B to C. Also a little geometry shows us that AB = BC. The time taken to move from C to D and then again from D to A is again same. The time period of the SHM of the shadow is castant.
Now let's check for it's acceleration at some point P after a time 't' has elapsed.

The horizontal component of acceleration at $Pis{a}_{Ch}$
$a_{Ch}=asin(\omega t+\frac{\pi }{2}-\theta )$
which is an acceleration relation for SHM.

Answer: Option B. -> 4J
:
B
$KE=\frac{1}{2}m(\omega \sqrt{A^2-x^2}{)}^2$
K.E. is max at x = 0 (mean position)
$\Rightarrow K{E}_{max}=\frac{1}{2}m\left({\omega }^2{A}^2\right)$
$=\frac{1}{2}2\left(2^2\right)$
$K{E}_{max}=4J.$

Answer: Option D. -> T12
:
D
Let the displacement of the particle be given by x = A sin $\omega$ t = A sin $\left(\frac{2\pi t}{T}\right)$ i.e., when x = 0, $t_0=0.$
When x = A/2, the value of t is given by $\frac{A}{2}=Asin\frac{2\pi t_1}{T};sin\frac{2\pi t_1}{T}=\frac{1}{2}$
$\frac{2\pi t_1}{T}=\frac{\pi }{6}or{t}_1=\frac{T}{12};t_1-t_0=\frac{T}{12}-0=\frac{T}{12}$