JEE Advanced · 2024 · 26 May 2024 · Paper 1 · Waves & Sound

Two uniform string of mass per unit length $\mu$ and $4\mu$, and length $L$ and $2L$, 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 $v_0=\frac{1}{2\mathrm{L}}\sqrt{\frac{\mathrm{T}}{\mu }}$, which of the following statement(s) is (are) correct ?

1. (A)With a node at O , the minimum frequency of vibration of the composite string is $v_0$.✓ Correct
2. (B)With an antinode at O , the minimum frequency of vibration of the composite string is $2v_0$.
3. (C)When the composite string vibrates at the minimum frequency with a node at O , it has 6 nodes, including the end nodes.✓ Correct
4. (D)No vibrational mode with an antinode at O is possible for the composite string.✓ Correct

Answer: (A), (C), (D)

Solution

Wave speeds: $C_1=\sqrt{\frac{T}{\mu }}$, $C_2=\sqrt{\frac{T}{4\mu }}=\frac{C_1}{2}$. For a node at O: $L=\frac{n{\lambda }_1}{2}$, $2L=\frac{m{\lambda }_2}{2}$. Since $\frac{C_1}{{\lambda }_1}=\frac{C_2}{{\lambda }_2}$, we get $m=4n$. Minimum frequency occurs at $n=1,m=4$: $v_{\min }=\frac{C_1}{2L}=v_0$ (A correct). The string has 1 loop in PO and 4 loops in OQ, giving 6 nodes total (C correct). For an antinode at O: $L=(2n+1)\frac{{\lambda }_1}{4}$, $2L=(2m+1)\frac{{\lambda }_2}{4}$. This requires $4=\frac{2m+1}{2n+1}$ (even = odd/odd), which is impossible (D correct). Option B is incorrect as no antinode mode exists.

JEE Advanced · 2024 · 26 May 2024 · Paper 1 · One or more correct options