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inessss [21]
3 years ago
14

(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

ve number of a period of 0.20 s, and an amplitude of 3.0 mm. Take the transverse direction to be the z direction.
(b) What is the maximum transverse speed of a point on the cord?
(c) What is the wave speed?
Physics
1 answer:
Vedmedyk [2.9K]3 years ago
7 0

Missing data: the wave number

k=60 cm^{-1}

(a)  z = 0.003 sin (6000y-31.4 t)

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

z = A sin (ky-\omega t)

where

A is the amplitude of the wave

k is the wave number

\omega is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

k = 60 cm^{-1} = 6000 m^{-1} is the wave number

T = 0.20 s is the period, so the angular frequency is

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s

So, the wave equation (in meters) is

z = 0.003 sin (6000y-31.4 t)

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

v_t = z' = -A \omega cos (ky-\omega t)

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

v_{t}_{max} =\omega A

where

\omega = 31.4 rad/s\\A = 0.003 m

Substituting,

v_{t}_{max}=(31.4)(0.003)=0.094 m/s

(c) 5.24 mm/s

The wave speed is given by

v=f \lambda

where

f is the frequency of the wave

\lambda is the wavelength

The frequency can be found from the angular frequency:

f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz

While the wavelength can be found from the wave number:

\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m

Therefore, the wave speed is

v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s

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Answer:

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