NTA Abhyas JEE Main2020PhysicsWaves and SoundPractice
A string of mass per unit length μ is clamped at both ends such that one end of the string is at x = 0 and the other is at x = L . When the string vibrates in fundamental mode, the amplitude of the mid-point O of the string is a , and tension in the string is T , Find the total oscillation energy stored in the string.
Options
- Aπ 2 a 2 T 4 L
- Bπ a T 2 L
- C- π 2 a 2 T 3 L
- Dπ 2 a T 6 L
Correct answer
A. π 2 a 2 T 4 L
Step-by-step solution
λ 2 = L λ = 2 L τ = μ V 2 The amplitude at a distance x from the origin is given by A = a sinkx consider an element of mass dm and length dx of string at a distance x from the origin The total energy of this element = its maximum KE = 1 2 d m ω 2 A 2 = 1 2 μ d x 4 π 2 f 2 a 2 sin 2 k x (Because μ = dm dx and ω = 2πf and it is given that A = a sinkx ) total energy of string = ∫ 0 L 2 π 2 μ f 2 a 2 sin 2 k x d x = π 2 μ f 2 a 2 [ x - sin 2 k x 2 ] 0 L = π 2 μ f 2 a 2 L = π 2 μ V 2 λ 2 a 2 L = π 2 4 L 2 T a 2