Question

In the given figure a massless rod of length $L$ suspended by two identical strings $PQ$ and $RS$ at the two ends. A block of mass $m$ is hung from point $O$. It is observed that the frequency of $2^{nd}$ harmonic in $PQ$ is equal to $3^{rd}$ harmonic frequency in $RS$. If $QO$ is equal to $\frac{4L}{x}$; then value of $x$ is ____.

Answer 1

13

Answer 2

1. The frequency of the $n^{th}$ harmonic of a stretched string is given by

   $$f_n = \frac{n}{2\ell}\sqrt{\frac{T}{\mu}},$$

   where $\ell$ is the length of the string, $T$ the tension, and $\mu$ the linear density.

2. For string $PQ$ (tension $T_1$), the 2nd harmonic frequency is

   $$f_2^{PQ} = \frac{2}{2\ell}\sqrt{\frac{T_1}{\mu}} = \frac{1}{\ell}\sqrt{\frac{T_1}{\mu}}.$$

   For string $RS$ (tension $T_2$), the 3rd harmonic frequency is

   $$f_3^{RS} = \frac{3}{2\ell}\sqrt{\frac{T_2}{\mu}}.$$

3. Given that

   $$f_2^{PQ} = f_3^{RS},$$

   we have

   $$\frac{1}{\ell}\sqrt{\frac{T_1}{\mu}} = \frac{3}{2\ell}\sqrt{\frac{T_2}{\mu}}.$$

   Canceling the common factors,

   $$\sqrt{T_1} = \frac{3}{2}\sqrt{T_2} \quad \Longrightarrow \quad T_1 = \frac{9}{4}\,T_2.$$

4. Now, consider the massless horizontal rod of length $L$ suspended by the two strings. Let the rod’s endpoints be $Q$ and $S$. A block of mass $m$ hangs from point $O$ (located on the rod) with $QO=\frac{4L}{x}$ and hence $OS = L - \frac{4L}{x}$.

5. For rotational equilibrium about $O$, the torques due to the tensions must balance:

   $$T_1 \times QO = T_2 \times OS.$$

   Substitute $T_1 = \frac{9}{4}\,T_2$ and $QO=\frac{4L}{x}$:

   $$\frac{9}{4}\,T_2 \cdot \frac{4L}{x} = T_2 \left(L - \frac{4L}{x}\right).$$

   Cancel $T_2$ (non-zero) and simplify:

   $$\frac{9L}{x} = L - \frac{4L}{x}.$$

   Multiply by $x$ to clear the denominator:

   $$9L = Lx - 4L.$$

   Dividing by $L$:

   $$9 = x - 4 \quad \Longrightarrow \quad x = 13.$$