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

Answer:

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

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

$U=\dfrac{dy}{dt}$

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

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

The maximum velocity

Umax = 1200 mm/s

Umax= 1.2 m/s