Question

In the figure shown, two identical blocks each of mass $m$ are connected to three massless springs of same natural length '$l_0$' each. If both the blocks are displaced slowly by $x_0$ each from their mean position and released, then the time period of oscillation of each block is $T_1 = 2\pi \sqrt{\frac{m}{aK}}$, when they are oscillating in opposite phase and $T_2 = 2\pi \sqrt{\frac{m}{bK}}$ when they are oscillating in same phase, then $a \times b$ = _______.

Answer 1

9

Answer 2

We have two masses $m$ connected by three springs: left and right springs each with spring constant $K$ and a middle spring with spring constant $4K$.

Step 1. Write the potential energy.

• Left spring: $U_L=\frac{1}{2}Kx_1^2$.
• Right spring: $U_R=\frac{1}{2}Kx_2^2$.
• Middle spring (extension $x_2-x_1$): $U_M=\frac{1}{2}(4K)(x_2-x_1)^2 =2K(x_2-x_1)^2$.

Step 2. Consider the two normal modes.

1. In-phase (symmetric) mode:
   Let $x_1=x_2=x$. Then

$$U_{\text{total}}=\frac{1}{2}Kx^2+\frac{1}{2}Kx^2 + 2K(0)^2 = Kx^2.$$

For each block the effective spring constant is $K$, so the angular frequency is

$$\omega_2 =\sqrt{\frac{K}{m}},$$

and hence

$$T_2=2\pi\sqrt{\frac{m}{K}}.$$

Comparing with given $T_2=2\pi\sqrt{\frac{m}{bK}}$, we have $b=1$.

2. Opposite-phase (antisymmetric) mode:
   Let $x_1=x$ and $x_2=-x$. Then the middle spring extension becomes

$$x_2-x_1= -x-x=-2x.$$

So,

$$U_M=2K(2x)^2=8Kx^2.$$

Now, the side springs contribute:

$$U_L=\frac{1}{2}Kx^2,\quad U_R=\frac{1}{2}Kx^2,$$

and the total potential energy is

$$U_{\text{total}} = \frac{1}{2}Kx^2+\frac{1}{2}Kx^2+8Kx^2 =9Kx^2.$$

The total kinetic energy (since both masses move with speed $\dot{x}$ but in opposite directions) is

$$T=\frac{1}{2}m\dot{x}^2+\frac{1}{2}m\dot{x}^2= m\dot{x}^2.$$

Therefore, the effective spring constant for the mode is $9K$ and the angular frequency is

$$\omega_1=\sqrt{\frac{9K}{m}}.$$

Thus,

$$T_1=2\pi\sqrt{\frac{m}{9K}}.$$

Comparing with $T_1=2\pi\sqrt{\frac{m}{aK}}$ we obtain $a=9$.

Step 3. Calculate:

$$a\times b=9\times 1=9.$$

Explanation (minimal):

1. For in-phase mode, only side springs act, so $b=1$.
2. For opposite-phase mode, energy gives effective constant $9K$, so $a=9$.
3. Thus, $a \times b = 9$.