**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 = 2π^{2}pn^{2}a^{2}

Energy transferred across a sq. cm per second= 2π^{2}pna^{2}v.

**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 y_{2 }=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

y_{1} = 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.