Question

The equation of a transverse wave traveling along a string is 1 1 y (2.00 mm)sin[(20 m )x (600 s )t] − − = − Find the (a) amplitude, (b) frequency, (c) velocity (including sign), and (d) wavelength of the wave. (e) Find the maximum transverse speed of a particle in the string

Answer 1

a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V= 30 m/s ( + x direction )

d) λ = 0.31 m

e)Umax= 1.2 m/s

Step-by-step explanation:

Given that

`y=2\ mm\ sin[(20m^(-1))x-(600s^(-1))t]`

As we know that standard form of wave equation given as

`y=A sin(\phi -\omega t)`

A= Amplitude

ω=Frequency (rad /s)

t=Time

Φ = Phase difference

`y=2\ mm\ sin[(20m^(-1))x-(600s^(-1))t]`

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s (2πf=ω)

ω= 2πf

f= ω /2π

f= 300/π = 95.49 Hz

K= 20 rad/m

So velocity,V

V= ω /K

V= 600 /20 = 30 m/s ( + x direction )

V = f λ

30 = 95.49 x λ

λ = 0.31 m

We know that speed is the rate of displacement

`U=(dy)/(dt)`

`U=2\ mm\ sin[(20m^(-1))x-(600s^(-1))t]`

`U=1200\ cos[(20m^(-1))x-(600s^(-1))t]\ mm/s`

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s