BT FT 6

At equilibrium, let x₀ be the spring elongation so that 2mg = kx₀.
If during the motion the body is at a position x (whose direction I choose to be downwards) from the equilibrium, then the total energy
-mgx +\dfrac{1}{2}mv^2 +\dfrac{1}{2}k\left(x_0+\dfrac{x}{2}\right)^2=\text{ constant}
Differentiating this equation w.r.t. time and using the fact that the velocity v = dxdt, we get finally the equation
\dfrac{\mathrm d^2x}{\mathrm dt^2}+\dfrac{k}{4m}\,x=0
from where we conclude that the angular frequency ω squared i.e. ω² = k4m
The FBD of mass m during motion is mg - T = ma
The maximum amplitude A_max is determined by the fact that the tension in the thread becomes zero when the maximum acceleration is obtained. i.e. when a = ω²A_max. Hence we get g = ω²A_max i.e.
A_\mathrm{max} =\dfrac{4mg}{k}.
Accordingly, α = 4.

Thanks a lot sir for those solutions.

Sir, for Q12. will option (a) be correct?

In Q10.

@Grandmaster,

that equation you have written is not valid for all instants. You just took the initial instant and the instant when the block passes through the equilibrium. That x appearing in your equation is not a variable; its just the initial displacement.

@kaymant sir: in 11, can u please elaborate on how (D) is also correct?

The initial momentum of the system (= plank + everything the plank contains) is zero. When the block (B) compresses the left spring maximally, the block must be stationary w.r.t. to the plank (P) and since the velocity of center of mass must remain zero, the speeds of B and P must be zero individually. Energy considerations give the compression of the left spring at this instant as \ell. As such the displacement (in magnitude) of B w.r.t. P is 5\ell

Since displacement of CM is zero, we get that
M\Delta \vec{r}_{PG}+m\Delta \vec{r}_{BG}=0
where \Delta \vec{r}_{PG} denotes the displacement of Plank P w.r.t. the ground G and similarly for the block B. But
\Delta \vec{r}_{BG}=\Delta \vec{r}_{BP}+\Delta\vec{r}_{PG}
Hence we get
\Delta \vec{r}_{PG}=-\dfrac{m}{m+M}\Delta \vec{r}_{BP}
Hence magnitude-wise, we get
|\Delta \vec{r}_{PG}|=\dfrac{m}{m+M}|\Delta \vec{r}_{BP}|=\dfrac{m}{m+M}\,5\ell
Hence (A).

Edit: (D) is not correct.
Since the rod just reaches the vertical position, its initial angular speed is obtained by
\dfrac{1}{2}\cdot\dfrac{1}{3}m\ell^2 \omega_0^2 = mg\dfrac{\ell}{2} which gives the initial angular speed as
\omega_0=\sqrt{\dfrac{3g}{\ell}}. Hence (B).
This angular speed comes from an impulsive torque which arises due to the magnetic force. If the current I(t) flows through the rod then the torque about the left end is
\tau = \int_0^\ell I(t)Bx\ \mathrm dx= I(t)B\dfrac{\ell^2}{2}
where I have measured the position x from the left end. The impulse of this toque is
\int_0^{\Delta t}\tau \ \mathrm dt = \dfrac{B\ell^2}{2}\int_0^{\Delta t} I(t)\ \mathrm dt= qB\dfrac{\ell^2}{2}
This impulse is responsible for the increment in angular momentum. Hence
qB\dfrac{\ell^2}{2}=\dfrac{1}{3}m\ell^2\omega_0
which gives
q=\dfrac{2m\omega_0}{3B}=\dfrac{2m}{3B}\sqrt{\dfrac{3g}{\ell}}
Hence (A) as well.

@decoder: how can C,D be simultaneously true?