JEE Advanced
Physics
Waves and Sound
2024
JEE Advanced 2024 (Paper 1)
JEE Advanced Physics Question (2024) — Solution
Question
Two uniform strings of mass per unit length and 4 , 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= 1 2 L T , which of the following statement(s) is(are) correct?
Options
- 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.
Answer
D. No vibrational mode with an antinode at O is possible for the composite string.
Step-by-step solution
C _1= T , C _2= T 4 = C _1 2 For node at O : L = n _1 2 , 2 ~L = m _2 2 ( n , m are integers) aligned & _1= 2 ~L n , _2= 4 ~L ~m \\ & C _1 _1 = C _2 _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 =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) _1 4 ; 2 ~L =(2 n +1) _2 4 (n, m are integers) \\ & _1= 4 ~L (2 n +1) ; _2= 8 ~L (2 ~m +1) \\ & C _1 _1 = C _2 _2 \\ & C _1 C _2 = _1 _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
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