Concepts Of Physics MCQ Edition [Volume 1]PhysicsWave Motion and Waves on a String
A wave travels along the positive x -direction with a speed of 20 m s ⁻¹ . The amplitude of the wave is 0.20 cm and the wavelength is 2.0 cm . Assuming the wave is described by the equation y = (0.20 cm ) [( cm ⁻¹)x - (2 10^3 s ⁻¹)t] , determine the displacement and velocity of the particle at x = 2.0 cm at time t = 0 .
Options
- ADisplacement is zero, velocity is -4 m s ⁻¹
- BDisplacement is zero, velocity is 4 m s ⁻¹
- CDisplacement is 0.20 cm , velocity is zero
- DDisplacement is -0.20 cm , velocity is zero
Correct answer
A. Displacement is zero, velocity is -4 m s ⁻¹
Step-by-step solution
The given wave equation is y = 0.20 ( x - 2 10^3 t) , where y and x are in cm and t is in s. Substituting x = 2.0 cm and t = 0 s into the equation: y = 0.20 ( 2.0 - 0) = 0.20 (2 ) = 0 The velocity of the particle is given by the partial derivative of displacement with respect to time: v_p = y t = t [0.20 ( x - 2 10^3 t)] v_p = 0.20 (-2 10^3) ( x - 2 10^3 t) v_p = -400 ( x - 2 10^3 t) cm s ⁻¹ Substituting x = 2.0 cm and t = 0 s: v_p = -400 (2 ) = -400 cm s ⁻¹ Converting to m s ⁻¹ : v_p = -4 m s ⁻¹ Thus, the displace