Answer: Option C. -> 1 s
:
C
$y=Asin\omega t$
$\frac{A}{2}=Asin\frac{2\pi }{12}t;\frac{1}{2}=sin\frac{\pi }{6}t\therefore \frac{\pi t}{6}=\frac{\pi }{6}\Rightarrow t=1s$

Answer: Option B. -> Periodic motion
:
B
You know that a motion that repeats itself in equal intervals of time is called periodic, right? Let's discuss what it means exactly. It means that after some specific interval of time T the position of the body from the reference point and its velocity are the same. So no matter what reference point you choose in the path, after time T the motion "repeats” itself. Well! Oscillatory motion is to and fro motion which may be periodic or not but the important thing about oscillatory motion is that it happens about a point of stable equilibrium where a restoring force exists. In the case mentioned in the problem there is no restoring force. However the motion repeats after equal intervals. So the motion is periodic but not Oscillatory!

Answer: Option D. -> 2 π/√α
:
D
In a simple harmonic motion $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$ where $\omega$ is the angular frequency. Comparing it with $\frac{d^{2}x}{d{t}^2}=-\alpha x$ we get ${\omega }^2=\alpha or\omega =\sqrt{\alpha }or\frac{2\pi }{T}=\sqrt{\alpha }orT=\frac{2\pi }{\sqrt{\alpha }}$, Hence the correct choice is (d)

Answer: Option B. -> B
:
B
We are given $x=A+Bsin\omega t$
We can see that the second term basically is the equation of an SHM with amplitude B. What's the first term doing there? Let's think about it. If it were just B sin ωt, it would be an SHM about the origin. Adding a constant to this entire thing would just shift the mean position from origin to x = A. Now the particle is doing SHM about x = A with an amplitude of B. Yes B! because amplitude is how far the particle goes from the mean position. The new means position is x = A but the particle's max displacement is still B from it!

Answer: Option C. -> T=2π√xg
:
C
If k is the force constant, we have
mg=kx
or $\frac{m}{k}=\frac{x}{g}$
$\therefore T=2\pi \sqrt{\frac{m}{k}}=2\pi \sqrt{\frac{x}{g}}$

Answer: Option B. -> 2
:
B
Among these given graphs we need to figure out which has maximum K.E. at 4 seconds. Not a big deal! Let's just think through. Since the three experiments involve the same spring block system so the spring constant remains same for all. Now only (1) and (2) are in their mean position at t = 4 and (3) is in extreme so we can rule out (3). In the mean position K.E = T.E = $\frac{1}{2}K{A}^2$. Among (1) and (2), (1) has higher value of A(amplitude)
which we can see from the height it's bump!. So (1) has the maximum K.E.
Alternative way of thinking could be
Since this is a displacement - time graph and $\frac{dx}{dt}$means velocity so the slope of this graph gives velocity.
More slope $\Rightarrow$ more velocity $\Rightarrow$ more K.E. Among (1) (2) and (3), (1) has the maximum slope (by observation) and so maximum K.E.

Answer: Option B. -> 18 J
:
B
Velocity of the body at time t is
$v=\frac{dx}{dt}=(6\times 100)cos(100t+\frac{\pi }{4})cm{s}^{-1}$
$\therefore V_{max}=600cm{s}^{-1}=6m{s}^{-1}$
Maximum KE $\frac{1}{2}m{v}_{max}^2=\frac{1}{2}\times 1\times (6{)}^2=18J$
Hence the correct choice is (b)

Answer: Option C. -> E1+E2+2√E1E2
:
C
$E_1=\frac{1}{2}m{\omega }^2{x}^{2}or\sqrt{E_1}=x\sqrt{\frac{1}{2}m{\omega }^2}.......\left(1\right)$
$E_2=\frac{1}{2}m{\omega }^2{y}^{2}or\sqrt{E_2}=y\sqrt{\frac{1}{2}m{\omega }^2}.......\left(2\right)$
$E=\frac{1}{2}m{\omega }^2(x+y{)}^{2}or\sqrt{E}=(x+y)\sqrt{\frac{1}{2}m{\omega }^2}.......(3)$
From (1), (2) and (3) it follows that
$\sqrt{E}=\sqrt{E_1}+\sqrt{E_2}$
$OrE=E_1+E_2+2\sqrt{E_1{E}_2}$
Which is choice (c).

Answer: Option D. -> 34
:
D
Kinetic energy $\left(KE\right)=\frac{1}{2}m{\omega }^2(A^2-x^2)$
Potential energy $\left(PE\right)=\frac{1}{2}m{\omega }^2{x}^2$
Total energy $\left(E\right)=\frac{1}{2}m{\omega }^2{A}^{2}Whenx=\frac{A}{2}$
KE = $\frac{1}{2}m{\omega }^2(A^2-\frac{A^2}{4})=\frac{3}{8}m{\omega }^2{A}^{2}E=\frac{1}{2}m{\omega }^2{A}^2\therefore \frac{KE}{E}=\frac{3}{4}$

Answer: Option C. -> T1=T2
:
C
Watch the next video for solution.