Question

A transverse, wave travelling on a chord is represented by D=0.22sin (5.6x+34t) where D and x are inmeters and t is in seconds. For this wave, determine; a) wavelength b) frequency c) velocity (both magnitude and direction) d) amplitude e) maximum and minimum speed of particles in the chord.

Answer 1

Answer:

a) λ = 1.12 m

b) f = 5.41 Hz

c) v = 154.54 m/s

d) A = 0.22m

e)

$v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}$

Explanation:

You have the following equation for a wave traveling on a cord:

$D=0.22sin(5.6x+34t)$     (1)

The general expression for a wave is given by:

$D=Asin(kx-\omega t)$    (2)

By comparing the equation (1) and (2) you have:

A: amplitude of the wave = 0.22m

k: wave number = 5.6 m^-1

w: angular velocity = 34 rad/s

a) The wavelength is given by substitution in the following expression:

$\lambda=\frac{2\pi}{k}=\frac{2\pi}{5.6m^{-1}}=1.12m$

b) The frequency is:

$f=\frac{\omega}{2\pi}=\frac{34s^{-1}}{2\pi}=5.41Hz$

c) The velocity of the wave is:

$v=\frac{\omega}{k}=\frac{34s^{-1}}{0.22m^{-1}}=154.54\frac{m}{s}$

d) The amplitude is 0.22m

e) To calculate the maximum and minimum speed of the particles you obtain the derivative of  the equation of the wave, in time:

$v_D=\frac{dD}{dt}=(0.22)(34)cos(5.6x+34t)\\\\v_D=7.48cos(5.6x+34t)$

cos function has a minimum value -1 and maximum +1. Then, you obtain for maximum and minimum velocity:

$v_D_{max}=7.48\frac{m}{s}\\\\v_D_{min}=-7.48\frac{m}{s}$