NTA Abhyas JEE Main2020PhysicsWaves and SoundPractice
A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end A as shown in the figure ( Total length of both strings combined is 2 L ). A block of mass M is tied to the free end of heavy string. Mass per unit length of the strings are μ and 16 μ and tension is T . Find lowest positive value of frequency such that junction point A is a node.
Options
- A1 L T μ
- B5 2 L T μ
- C3 2 L T μ
- D1 2 L T μ
Correct answer
D. 1 2 L T μ
Step-by-step solution
f 1 = n 1 2 L T μ , f 2 = n 2 2 L T 16 μ f 1 = f 2 ⇒ n 1 = n 2 4 n 1 = 1 , n 2 = 4 f min = 1 2 L T μ