Concepts Of Physics MCQ Edition [Volume 1]PhysicsWave Motion and Waves on a String
A string clamped at both ends is set into vibration by a 660 Hz tuning fork, causing it to vibrate in three loops. The wave speed for a transverse wave on the string is 220 m s ⁻¹ . If the maximum amplitude of a particle is 0.5 cm , which of the following is a suitable equation describing the motion?
Options
- A(0.5 cm ) [(0.06 cm ⁻¹)x] [(1320 s ⁻¹)t]
- B(0.5 cm ) [(0.06 cm ⁻¹)x] [(660 s ⁻¹)t]
- C(0.5 cm ) [(0.06 cm ⁻¹)x] [(1320 s ⁻¹)t]
- D(0.5 cm ) [(6 cm ⁻¹)x] [(1320 s ⁻¹)t]
Correct answer
A. (0.5 cm ) [(0.06 cm ⁻¹)x] [(1320 s ⁻¹)t]
Step-by-step solution
Given frequency f = 660 Hz and wave speed v = 220 m s ⁻¹ = 22000 cm s ⁻¹ . Angular frequency = 2 f = 2 660 = 1320 rad s ⁻¹ . Wave number k = v = 1320 22000 = 0.06 cm ⁻¹ . Since the string is clamped at both ends, the displacement at x = 0 must be zero at all times. Therefore, the spatial part of the standing wave equation must be a sine function, (kx) . The maximum amplitude of the standing wave is given as A = 0.5 cm . The equation of the standing wave is of the form y(x,t) = A (kx) ( t) . Substituting the values,