Question 47: Waves

Question

Two uniform strings of mass per unit length $\mu$ and $4 \mu$, and length $L$ and $2 L$, respectively, are joined at point $\mathrm{O}$, and tied at two fixed ends $\mathrm{P}$ and $\mathrm{Q}$, as shown in the figure. The strings are under a uniform tension $T$. If we define the frequency $v_0=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$, which of the following statement(s) is(are) correct?

Accepted Answer

$$C_1=\sqrt{\frac{T}{\mu}}, C_2=\sqrt{\frac{T}{4 \mu}}=\frac{C_1}{2}$$
For node at O :
$$\begin{aligned}
& \mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}(\mathrm{n}, \mathrm{m} 	ext { are integers }) \
& \lambda_1=\frac{2 \mathrm{~L}}{\mathrm{n}}, \lambda_2=\frac{4 \mathrm{~L}}{\mathrm{~m}} \
& \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \
& \Rightarrow \frac{\mathrm{C}_1}{\frac{2 \mathrm{~L}}{\mathrm{n}}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{2 \mathrm{~L}}{\mathrm{~m}}} \
& \Rightarrow 4 \mathrm{n}=\mathrm{m}
\end{aligned}$$
For minimum frequency, $$\mathrm{n}=1, \mathrm{~m}=4$$
$$	herefore v_{\min }=\frac{\mathrm{C}_1 	imes 1}{2 \mathrm{~L}}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}=v_0$$
The string will look like
Total no. of nodes $$=6$$ including the end nodes
For antinode at O :
$$\begin{aligned}
& \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad 	ext { ( } \mathrm{n}, \mathrm{m} 	ext { are integers) } \
& \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \
& \frac{C_1}{\lambda_1}=\frac{C_2}{\lambda_2} \
& \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \
& 2=\frac{\frac{4 L}{(2 n+1)}}{\frac{8 \mathrm{~L}}{(2 m+1)}} \
\end{aligned}$$
$$4=\frac{(2 m+1)}{(2 n+1)} \Rightarrow 	ext { even }=\frac{	ext { odd }}{	ext { odd }} \Rightarrow 	ext { This node is not possible }$$

Answer

With a node at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $v_0$.

Answer

With an antinode at $\mathrm{O}$, the minimum frequency of vibration of the composite string is $2 v_0$.

Answer

When the composite string vibrates at the minimum frequency with a node at $\mathrm{O}$, it has 6 nodes, including the end nodes.

Answer

No vibrational mode with an antinode at $\mathrm{O}$ is possible for the composite string.

Waves

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