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Two uniform strings of mass per unit length $\mu$ and $4 \mu$, and length $L$ and $2 L$, respectively, are joined at point $O$, and tied at two fixed ends $P$ and $Q$, 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?
$(A)$ With a node at $O$, the minimum frequency of vibration of the composite string is $v_0$
$(B)$ With an antinode at $O$, the minimum frequency of vibration of the composite string is $2 v_0$
$(C)$ When the composite string vibrates at the minimum frequency with a node at $O$, it has $6$ nodes, including the end nodes
$(D)$ No vibrational mode with an antinode at $O$ is possible for the composite string
$A,C,D$
$A,C$
$A,B,C$
$A,B,D$
Solution
(image)
$C _1=\sqrt{\frac{ T }{\mu}}, C _2=\sqrt{\frac{ T }{4 \mu}}=\frac{ C _1}{2}$
For node at $O$ :
$L =\frac{ n \lambda_1}{2}, 2 L =\frac{ m \lambda_2}{2} \text { (n, } m \text { are integers) }$
$\lambda_1=\frac{2 L }{ n }, \lambda_2=\frac{4 L }{ m }$
$\frac{ C _1}{\lambda_1}=\frac{ C _2}{\lambda_2}$
$\Rightarrow \frac{ C _1}{\frac{2 L }{ n }}=\frac{\frac{ C _1}{2}}{\frac{4 L }{ m }}$
$\Rightarrow 4 n = m$
For minimum frequency, $n =1, m =4$
$\therefore v_{\min }=\frac{ C _1 \times 1}{2 L }=\frac{1}{2 L } \sqrt{\frac{ T }{\mu}}=v_0$
The string will look like
(image)
Total no. of nodes $=6$ including the end nodes
For antinode at $O$ :
$L =(2 n +1) \frac{\lambda_1}{4} ; 2 L =(2 n +1) \frac{\lambda_2}{4} \quad \text { (n, } m \text { are integers) }$
$\lambda_1=\frac{4 L }{(2 n +1)} ; \lambda_2=\frac{8 L }{(2 m +1)}$
$\frac{ C _1}{\lambda_1}=\frac{ C _2}{\lambda_2}$
$\frac{ C _1}{ C _2}=\frac{\lambda_1}{\lambda_2}$
$2=\frac{\frac{4 L }{(2 n +1)}}{\frac{8 L }{(2 m +1)}}$
$4=\frac{(2 m +1)}{(2 n +1)} \Rightarrow \text { even }=\frac{\text { odd }}{\text { odd }} \Rightarrow \text { This node is not possible }$
