Two uniform strings of mass per unit length μ and 4 μ , 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

Correct answer: A. With a node at O , the minimum frequency of vibration of the composite string is v_0 . C_1= T μ , C_2= T 4 μ = C_1 2 For node at O : L= n λ_1 2 , 2 ~L= m λ_2 2 ( n, m are integers) aligned & λ_1= 2 ~L n , λ_2= 4 ~L ~m & C_1 λ_1 = C_2 λ_2 & C_1 2 ~L n = C_1 2 4 ~L m & 4 n=m aligned For minimum frequency, n=1, ~m=4 v_ = C_1 × 1 2 ~L = 1 2 ~L T μ =v_0 The string will look like Total no. of nodes =6 including the end nodes For antinode at O : aligned & L=(2 n+1) (λ_1)/(4) ; 2 ~L=(2 n+1) (λ_2)/(4) (n, m are integers) & λ_1= 4 ~L (2 n+1) ; λ_2= 8 ~L (2 ~m+1) & C_1 λ_1 = C_2 λ_2 & C_1 C_2 =(λ_1)/(λ_2) & 2= 4 ~L (2 n+1) 8 ~L (2 ~m+1) & 4= (2 ~m+1) (2 n+1) even = odd odd This node is not possible aligned

JEE Advanced 2024Paper 1

Exam: JEE Advanced
Year: 2024
Paper: Paper 1
Subject: Physics
Chapter: Waves and Sound

JEE Advanced 2024 · Paper 1 · Physics · Waves and Sound

Two uniform strings of mass per unit length μ and 4 μ , and length...

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? Two uniform strings of mass per unit length μ and 4 μ , and length L and 2 L , respectively, are joined at point O ,

Two uniform strings of mass per unit length μ and 4 μ , and length L and 2 L , respectively, are joined at point O ,

Options

AWith a node at $\mathrm{O}$ , the minimum frequency of vibration of the composite string is $v_0$ .
BWith an antinode at $\mathrm{O}$ , the minimum frequency of vibration of the composite string is $2 v_0$ .
CWhen the composite string vibrates at the minimum frequency with a node at $\mathrm{O}$ , it has 6 nodes, including the end nodes.
DNo vibrational mode with an antinode at $\mathrm{O}$ is possible for the composite string.

Correct answer

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

Step-by-step solution

\mathrm{C}_1=\sqrt{\frac{\mathrm{T}}{\mu}}, \mathrm{C}_2=\sqrt{\frac{\mathrm{T}}{4 \mu}}=\frac{\mathrm{C}_1}{2} \underline{\text{

For

node at } \mathrm{O} \text{ : }} \mathrm{L}=\frac{\mathrm{n} \lambda_1}{2}, 2 \mathrm{~L}=\frac{\mathrm{m} \lambda_2}{2}

(

\mathrm{n}, \mathrm{m}

are integers)

\begin{aligned} & \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}}{n}}=\frac{\frac{\mathrm{C}_1}{2}}{\frac{4 \mathrm{~L}}{m}} \\ & \Rightarrow 4 \mathrm{n}=\mathrm{m}\end{aligned}

For minimum frequency,

\mathrm{n}=1, \mathrm{~m}=4 \therefore v_{\min }=\frac{\mathrm{C}_1 \times 1}{2 \mathrm{~L}}=\frac{1}{2 \mathrm{~L}} \sqrt{\frac{\mathrm{T}}{\mu}}=v_0

The string will look like Two uniform strings of mass per unit length μ and 4 μ , and length L and 2 L , respectively, are joined at point O ,

Total no. of nodes

=6

including the end nodes

\underline{\text{ For antinode at } \mathrm{O} \text{ : }}

\begin{aligned} & \mathrm{L}=(2 \mathrm{n}+1) \frac{\lambda_1}{4} ; 2 \mathrm{~L}=(2 \mathrm{n}+1) \frac{\lambda_2}{4} \quad \text{ (n, m are integers) } \\ & \lambda_1=\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)} ; \lambda_2=\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)} \\ & \frac{\mathrm{C}_1}{\lambda_1}=\frac{\mathrm{C}_2}{\lambda_2} \\ & \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{\lambda_1}{\lambda_2} \\ & 2=\frac{\frac{4 \mathrm{~L}}{(2 \mathrm{n}+1)}}{\frac{8 \mathrm{~L}}{(2 \mathrm{~m}+1)}} \\ & 4=\frac{(2 \mathrm{~m}+1)}{(2 \mathrm{n}+1)} \Rightarrow \text{ even }=\frac{\text{ odd }}{\text{ odd }} \Rightarrow \text{ This node is not possible }\end{aligned}

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