Two uniform strings of mass per unit length m | vedclass

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(A,C,D) Let the wave speeds in the two strings be $v_1 = \sqrt{\frac{T}{\mu}}$ and $v_2 = \sqrt{\frac{T}{4\mu}} = \frac{v_1}{2}$.For a node at $O$:$L

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$A, C, D$

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(A,C,D) Let the wave speeds in the two strings be $v_1 = \sqrt{\frac{T}{\mu}}$ and $v_2 = \sqrt{\frac{T}{4\mu}} = \frac{v_1}{2}$.For a node at $O$:$L = \frac{n \lambda_1}{2}$ and $2L = \frac{m \lambda_2}{2}$,where $n, m$ are integers.Since the frequency $f = \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}$,we have $\frac{v_1}{2L/n} = \frac{v_1/2}{4L/m} \Rightarrow \frac{n v_1}{2L} = \frac{m v_1}{8L} \Rightarrow 4n = m$.For the minimum frequency,$n=1$,so $m=4$. The frequency is $f = \frac{v_1}{2L} = v_0$. Thus,$(A)$ is correct.For $n=1$ and $m=4$,the number of loops in the first string is $n=1$ (nodes at $P$ and $O$) and in the second string is $m=4$ (nodes at $O$ and $Q$).Total nodes = (nodes in string $1$) + (nodes in string $2$) - $1$ (common node at $O$) = $(n+1) + (m+1) - 1 = 2 + 5 - 1 = 6$. Thus,$(C)$ is correct.For an antinode at $O$:$L = (2n-1) \frac{\lambda_1}{4}$ and $2L = (2m-1) \frac{\lambda_2}{4}$.Frequency $f = \frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2} \Rightarrow \frac{v_1}{4L/(2n-1)} = \frac{v_1/2}{8L/(2m-1)} \Rightarrow \frac{2n-1}{4L} = \frac{2m-1}{16L} \Rightarrow 4(2n-1) = 2m-1$.Since $4(2n-1)$ is even and $2m-1$ is odd,this equation has no integer solution. Thus,$(D)$ is correct.

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