Two uniform strings of mass per unit length and 4 , 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 0= 12L T , which of the following statement(s) is(are) correct? [figure]
- With a node at O, the minimum frequency of vibration of the composite string is 0.
- With an antinode at O, the minimum frequency of vibration of the composite string is 2 0.
- When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes.
- No vibrational mode with an antinode at O is possible for the composite string.
Answer
(1) With a node at O, the minimum frequency of vibration of the composite string is 0., (3) When the composite string vibrates at the minimum frequency with a node at O, it has 6 nodes, including the end nodes., (4) No vibrational mode with an antinode at O is possible for the composite string.
Solution
Wave speeds: v1= T/ , v2= T/4 = v12.
Node at O (both parts fixed–fixed): f= p v12L=p 0 and f= q v22(2L)= q 04. Minimum common value: p=1, q=4 ⇒ f = 0 → (A) correct.
Then PO has 1 loop and OQ has 4 loops: nodes P, O, 3 inside OQ, Q = 6 nodes → (C) correct.
Antinode at O (both parts fixed–free): f= (2p-1)v14L= (2p-1) 02 and f= (2q-1)v24(2L)= (2q-1) 08.
Equality needs 4(2p-1)=2q-1: even = odd, impossible ⇒ no such mode → (D) correct, (B) wrong.
Answer: (A), (C), (D)