Just hanging around- part 1

Let $x$ be the (downward) displacement of the pulley (from equilibrium). Then the upper spring is extended by $2x$ , creating a tension of $2kx$ in the rope. Since this tension acts on both sides of the pulley, we have an upward force of $4kx$ on the pulley.

Now let $y$ be the (downward) displacement of the block of mass $m$ . Then the lower spring is extended by $y-x$ , and it will exert a downward force of $2k(y-x)$ on the pulley. Since the net force on the pulley must be zero, we have $4kx=2k(y-x)$ , or, $x=\frac{y}{3}$ . Now the lower spring exerts an upward force of $F=2k(y-x)=\frac{4}{3}ky$ on the block, meaning that the effective spring constant for the whole system is $k_{eff}=\frac{F}{y}=\frac{4}{3}k$ .

Finally the period is $T=2\pi\sqrt{\frac{m}{k_{eff}}}$ $=2\pi\sqrt{\frac{3m}{4k}},$ with $a=3$ and $b=4$ .

Let extension in the top spring be $y_1$ and the extension of the lower spring be $y_2$ .

Now, the tension is uniform throughout the string, and the tension $T$ , pulls at both points of contact with the ceiling, thus for the balance of forces on the massless string, we have

$T - ky_1 = 0$

For the balance of forces on the massless pulley, we have

$\begin{aligned} 0a &= 2T - 2ky_2 \\ &= 2ky_1 - 2ky_2 \end{aligned}$

which implies $y_1=y_2$

The total displacement for the rectangular block is given by $y=\frac{y_1}{2}+y_2 = \frac{3y_2}{2}$

Now for the block, we have

$\begin{aligned} m\frac{d^2y}{dt^2} &= mg - 2ky_2 \\ &= mg - \frac{4}{3}ky \end{aligned}$

If we perform the change of variables, $y \rightarrow y_{eq} + \Delta y$ , where $y_{eq} = \frac{3mg}{4k}$ , we have

$m\frac{d^2y}{dt^2} = -\frac{4}{3}ky$

which is the usual equation for simple harmonic motion, and we find

$\omega =\sqrt{\frac{4k}{3m}}$

$a=3,b=4$

$\sqrt{a^2+b^2}$ =5

It can also be done by writing mechanical energy equation and since no external force other than gravity acts therefore differentiate that equation and equate to 0.

Suppose we see the system when it is not in equilibrium. The two springs having displacements x(1) &x (2). The mass m applies force F=mw^2x on the spring which causes displacement x (1)= F/2k. Since the spring is masssless( gravity free condition) it pulls the pulley with the force F and ultimately force F/2 is transmitted to the 2nd spring x (2)=F/2k. The total dispacement is x (1)+x (2)/2 = 3F/4k. Finally F=mw^2 (3F/4k). w= 2♡(pi) (3m/4k)^1/2. a=3 b= 4.

Suppose we see the system when it is not in equilibrium. The two springs having displacements x(1) &x (2). The mass m applies force F=mw^2x on the spring which causes displacement x (1)= F/2k. Since the spring is masssless( gravity free condition) it pulls the pulley with the force F and ultimately force F/2 is transmitted to the 2nd spring x (2)=F/2k. The total dispacement is x (1)+x (2)/2 = 3F/4k. Finally F=mw^2 (3F/4k). w= 2♡(pi) (3m/4k)^1/2. a=3 b= 4.