Concepts Of Physics MCQ Edition [Volume 1]PhysicsWave Motion and Waves on a String
A string, 2 m long and fixed at both ends, is set into vibrations in its first overtone. The wave speed on the string is 200 m s ⁻¹ and the amplitude is 0.5 cm . Determine the equation giving the displacement of different points as a function of time. Choose the X -axis along the string with the origin at one end and t = 0 at the instant when the point x = 50 cm has reached its maximum displacement.
Options
- A(0.5 cm ) [(2 m ⁻¹)x] [(100 s ⁻¹)t]
- B(0.5 cm ) [ ( 2 m ⁻¹ )x] [(100 s ⁻¹)t]
- C(0.5 cm ) [( m ⁻¹)x] [(200 s ⁻¹)t]
- D(0.5 cm ) [( m ⁻¹)x] [(200 s ⁻¹)t]
Correct answer
D. (0.5 cm ) [( m ⁻¹)x] [(200 s ⁻¹)t]
Step-by-step solution
For a string fixed at both ends, the wavelength in the first overtone ( n = 2 ) is given by L = 2 ( 2 ) = . Given L = 2 m , we have = 2 m . The wave number k is k = 2 = m ⁻¹ . The angular frequency is = v k = 200 = 200 s ⁻¹ . The general equation for a standing wave with nodes at x = 0 and x = L is y(x, t) = A (kx) ( t + ) . Given the amplitude A = 0.5 cm . At t = 0 , the point x = 50 cm = 0.5 m has maximum displacement. Substituting x = 0.5 m into the spatial part gives ( 0.5) = 1 , which means it is an antinode.