Question

The displacement of a wave traveling in the positive x-direction is D (, t)(3.5cm)si (3.2c - 132t), where c is in m and t is in s. You may want to review (Pages 427 - 430)

A) What is the frequency of this wave?

B) What is the wavelength of this wave?

C) What is the speed of this wave?

Answer 1

given

D ( x,t ) = 3.5 cm sin ( 3.2x - 132t ) ------- 1

using this above equation D ( x,t ) = A sin ( k x + ω t ) ---------- 2

comparing equation 1 and 2 we get the following results

the amplitude of the wave is A = 3.5 cm

the wave number of the wave is k = 3.2

2$\pi$ / $\lambda$ = 3.2

$\lambda$ = 2$\pi$ / 3.2

= 6.28 / 3.2

$\lambda$ = 1.9625 m

the frequency of the wave is f = ω / 2$\pi$

= 132 / 2 x 3.14

f = 21.0191 Hz

and the speed of the wave is V = $\lambda$ f

= 1.9625 x 21.0191

V = 41.25 m/sec

so the final answers are :-

A ) the frequency of this wave is f = 21.0191 Hz

B ) the wavelength of this wave is $\lambda$ = 1.9625 m

C ) the speed of this wave is V = 41.25 m/sec