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Numericals Based on Progressive Waves

 

Problem 1. When a simple harmonic wave is propagated through a medium, the displacement of a particle (in cm) at any instant of time is given by

 

y = 10 sin 2π /100 (3600 t – 20)

 

Calculate, the amplitude of vibrating particle, wave velocity, wavelength, frequency and time period.

 

Hint:                Y=a sin 2π /λ (vt –x).

 

N= v/λ ,t =1/n

 

[Ans. a = 10 cm, A.= 100 cm, v = 3600 cm/sec, n =360Hz, T =1/360 second.]

 

 

Problem 2.   A source of sound has a frequency of 512 Hz and an amplitude of 0.25 cm. What is flow of energy across a square em per second, if the velocity of sound in air is 340 m/s and the density of air is 0.00129 glee?

 

Hint: Total energy per unit volume = 2π2pn2a2

Energy transferred across a sq. cm per second= 2π2pna2v.

[Ans. 1.417 x 107 erg / em2 8]

 

Problem 3.    The equation of stationary wave formed by superposition of two identical  progressive waves travelling in opposite direction is given by, y =A cos  kx sin rot, where           k = 2π /λ ,A= 1.0 mm, k= 1.57 cm-1 and w = 78.5 rad/sec. Calculate (a) velocity of superposing wave, (b) amplitude of vibration

of the wave at x = 2.33 cm.

 

Solution: The equation of the superposing waves are

 

y1 = A/ 2 sin (wt -kx) and y2 =A/2

 

Then (a) velocity v = w/k = = ( 50 m/ sec, )

 

(b) Amplitude a = A cos  kx = ( 0.866 mm.)

 

Problem 4. A simple harmonic wave of amplitude 8 units transverses a line of particles in the direction of the positive X-axis. At any given instant of time, for a particle at a distance of 10 cm from the origin, the displacement is +6 units and for a particle at a distance of 25 cm from the origin, the displacement is + 4 units. Calculate the wavelength.

 

Solution:  y= a sin 2π /λ (vt –x)

 

 

.

 

Problem 5. Show that the equation y = 4 cos 2π ( t/0.02-x/ 400) cm represents a progressive ware. Calculate its amplitude, wavelength, wave velocity and frequency.

 

          Solution: y = 4 cos 2π (t /0.02 –x/ 400) compare with y =a cos 2π(t/T- x/X) a  = 4, λ= 400 cm,  v 200m/sec ,n =v/λ =20000 /400 =50 Hz

 

At (t + 1) sec, displacement at distance (x + 20000) cm is

 

y1 = 4 cos  2π /400  [20000 (t + 1)- (x + 20000)]

 

= 4 cos 2π /400  ( 20000 t- x)

Y = Y  t

 

. . After 1 second at a distance 20000 em the displacement is same. Hence the given

equation represents a progressive wave.