1. A wave has the equation (x in metres and t in seconds) Y = 0.02Sin(30t – 4x). Find

1. Its frequency, speed and wave length.
2. The equation of wave with double amplitude but travelling in the opposite direction.
   Solution:
   Given,
   i.    Equation of wave
   y = 0.02sin(30t – 4x)
   Comparing it with general eq^n. of motion,
   y = [latex]asin(\omega t – kx[/latex]), we get,
   Amplitude of wave (a) = 0.02 m
   Angular frequency ([latex]\omega[/latex]) = 30
   Wave number (k) = 4 m^-1
   We know,
   [latex]\omega = 2\pi f[/latex]
   
   Or, f = [latex]\frac{\omega}{2\pi}[/latex]
   
              [latex]= \frac{30}{2\times 3.14} = 4.777 m[/latex]
   
   Now, k = [latex]\frac{2\pi}{\lambda}[/latex]
   
   Or, [latex]\lambda = \frac{2\pi}{k}[/latex]
   
             = [latex]\frac{2\times 3.14}{4}[/latex] = 1.57 m
   Again,
   Speed (v) = f.[latex]\lambda[/latex]
   
              = 4.777 x 1.57
              = 7.5 m/s
   ii.   Now, new amplitude (a’) = 2a
                                     = 2 x 0.02
                                     = 0.04
   So, equation of wave with double amplitude but travelling in opposite direction is
   y = [latex]a’sin(\omega t + kx)[/latex]
   
               = 0.04sin(30t + 4x).

2. A sound wave of frequency 400 Hz is travelling in air at a speed of 320 ms^-1. Calculate the difference in phase between two points on the wave 0.2 m apart in the direction of travel. Ans:1.57 radian
   Solution:
   Frequency (f) = 400 Hz
   Speed (v) = 320
   Phase difference ([latex]\Delta \phi[/latex]) =?
   Path difference ([latex]\Delta x[/latex]) = 0.2 m
   We know that,
   [latex]\Delta \phi = \frac{2\pi}{\lambda}\times \Delta x[/latex]
   
               = [latex]\frac{2\pi}{\frac{v}{f}}\times \Delta x   [\because \lambda = \frac{v}{f}[/latex]]
   
               = [latex]\frac{2\pi}{\frac{320}{400}}\times 0.2[/latex]
   
               = 1.57 radian
3. The equation of motion of a wave is y = 1.2sin(3.5t – 0.5x), where distance and time are expressed in meter and second respectively. Determine the amplitude, frequency, wavelength and velocity of the wave. Also, find the maximum speed of particles in that medium and the equation of a wave with double amplitude and frequency but travelling exactly in the opposite direction of the given wave. (Ans: 1.2m, 0.56 Hz, 12.57 m, 7.04 ms^-1, 4.2 ms^-1, 2.4sin(7.04t+0.5x)
   Solution:
   Equation of motion of a wave: y = 1.2sin(3.5t – 0.5x)
   Amplitude (a) =?
   Frequency (f) =?
   Wavelength ([latex]\lambda[/latex]) =?
   Velocity of wave (v) =?
   Comparing above eq^n. with general equation of progressive wave,
   [latex]y = asin(\omega t – kx)[/latex], we get,
   Amplitude (a) = 1.2 m
   [latex]\omega = 3.5[/latex]
   
   Or, [latex]2\pi f = 3.5[/latex]
   
   Or, f = [latex]\frac{3.5}{2\pi} = 0.557\ Hz[/latex]
   
   Also, k = 0.5
   
   Or, [latex]\frac{2\pi}{\lambda} = 0.5[/latex]
   
   Or, [latex]\lambda = \frac{2\pi}{0.5} = 12.566\ m[/latex]
   
   Now, wave – velocity (v) = [latex]f.\lambda[/latex]
   
                           = [latex]0.557\times 12.566 = 7\ ms^{-1}[/latex]
   
   For maximum speed of particle, we use,
   [latex]v = \omega a = 3.5\times 1.2 = 4.2\ ms^{-1}[/latex]
   
   Now, for equation of a wave with double amplitude and frequency but travelling exactly in the opposite direction of the given wave
   New amplitude (a’) = 2a = [latex]2\times 1.2 = 2.4\ m[/latex]
   
   New frequency (f’) = 2f = [latex]2\times 0.557[/latex] = 1.114 Hz
   
   So, Equation of required wave is: [latex]y = a’sin(\omega ‘ t + kx)[/latex]         [Positive sign is due to opposite direction]
   y = 2.4 sin ([latex]2\pi f’ t + kx)[/latex]
   
               = [latex]2.4sin(2\pi\times 1.114t + 0.5x)[/latex]
   
                                                   = [latex]2.4sin(7t + 0.5x)[/latex]