A Classical Approach of Newtonian Mechanics

13 WAVE MOTION 13.6 The Doppler effect

×

μ

The propagation speed of waves on the wire is given by

v = f λ = 261.6 × 1.80 = 470.88 m/s.

Furthermore, the string’s mass per unit length is

m

μ = =

L

5.4 10 −^3 − 3

0.9

= 6.00 × 10

kg/m.

Now, the relationship between the wave propagation speed, v, the mass per unit

length,^ μ,^ and^ the^ tension,^ T,^ of^ a^ stretched^ wire^ is^

v =

‚

., T

.

Thus,

T = v^2 μ = (470.88)^2 × 6.00 × 10 −^3 = 1.330 × 103 N.

Worked example 13.3: Sinusoidal wave

Question: A wave is described by

y = A sin (k x − ω t),

where A = 4 cm, k = 2.65 rad./m, and ω = 4.78 rad./s. Moreover, x is in meters

and t is in seconds. What are the wavelength, frequency, and propagation speed

of the wave?

Answer: We identify A as the wave amplitude, k as the wavenumber, and ω as

the angular frequency. Now, k = 2 π/λ, where λ is the wavelength. Hence,

λ =

2 π

=

2 × π

k 2.65

= 2.371 m.

Furthermore, ω = 2 π f, where f is the frequency. Hence,

ω

f = =

2 π

4.78 (^)
2 × π
= 0.7608 Hz.
Finally, v = f λ, where v is the propagation speed. Thus,
v = 0.7608 × 2.371 = 1.804 m/s.