Question

A wave traveling down a thin rope has a transverse displacement given by y(x,t)=(4.50mm)cos(218xm^(-1)+1550ts^(-1))

The rope has a length of 1.61m and a mass of 0.00425kg.

Determine the following:

(a) amplitude of the wave,

(b) frequency of the wave,

(c) wavelength of the wave,

(d) wavespeed,

(e) direction the wave is traveling,

(f) tension inthe rope

Answer 1

general eqn of wave is Y (x,t) = A cos (Kx-wt)

where A is Amplitude

K is waveno. = 2pi/L

where L is waveelngth

W   = 2pif

where f is frequency

so

now Comparing given eqn with this

a. A = 4.5 mm
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b.W = 2pi f = 1550

frequency f = 1550/2pi

f = 246.8 Hz

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c. K = 2pi/L = 218

wavelength L = 2pi/218

L = 0.0288 m

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speed V = Lf

V = 246.8 * 0.0288

V = 7.1 m/s

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+ x direction

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use the formula for tension T = V^2 u

where u is mass per unit length

so

u = 0.00425/1.61

u = 0.00263 kg/m

so

T = 7.1* 7.1 * 0.00263

T = 0.132 N