Question

The equation of a particular wave travelling along a wire is given as

y=(0.47m)sin[(150s−1)t+(53m−1)x]y=(0.47m)sin⁡[(150s−1)t+(53m−1)x]

1. Based on this equation, find the amplitude, wavelength, frequency, and velocity of the wave. Be sure to include the correct sign for the velocity.

   Amplitude = m Wavelength = m Frequency = Hz Velocity = m/s

2. What is the displacement of the wire at x=0.29x=0.29 m and t=0.26t=0.26 s?

   Displacement = m

I got the first three right but can't figure out how to get the last two.

The equation of a particular wave travelling along a wire is given as y= (0.47m) sin((150s-?)t + (53m-?)x] a. Based on this equation, find the amplitude, wavelength, frequency, and velocity of the wave. Be sure to include the correct sign for the velocity. Amplitude = 0.47 Wavelength = 0.119 m Frequency = 23.873 Hz Velocity = 2.830188 30188 m/s b. What is the displacement of the wire at x = 0.29 m and t = 0.26 s? Displacement = 0.382014 m

Answer 1

We have given

y = (0.47m) sin((150s-?)t + (53m-1).

Standard equation for the wave traveling in negative direction with velocity (v) may be written as

y = (A) sin((w)t + (k).

So on comparing

Amplitude will be A= 0.47m

Propagation vector k = 53 $m^{-1}$

b) in order to calculate the displacement put the variables values of x and t, then just solve it.  

6) Propagation vecty k = 53mt k= 211 x = 21 = 20 x 0.19 m Angulars frequency w = 150 W2200 v (frequenry) = 6 - 1 7 23:87 HZ Velocity (V) = ox (product of frequary & wavelength) = 23.87X0.119 = 284 m/s 6) Displacement of wire at a=0.29m and t=0.26 sec 4= (0.47m) Sim [ (150 x4)t +(3364) m] = (0.47m) #riſ (150 X 0-24) + (53X0-27)] + (017) 3im ( 54:37] ~ 0.382 m

* one important thing to note that sign of velocity will be negative (traveling in opposite directions).  

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