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Old Jun 27th 2009, 01:32 PM   #1
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Finding Amplitude of a Wave

A transverse sinusoidal wave on a string has a period of T= 25.0ms and travels in the negative x direction with a speed 30 m/s. At t=0, an element of the string at x=0 has a transverse position of 2 cm and is traveling downward at a speed of 2 m/s.

A) What is the amplification of the wave?
B) What is the initial phase angle?

For part A, given that I don't know the phase angle, I decided to try solving using the transverse velocity v=-wAcos(kx -wt)
So, -0.02m/s= -(80pi)Acos(0)
A=0.0000796 m

However, this answer is totally wrong. The book states A = 0.0215m.
W/out phase constant I don't know any other way to solve this. Help please?

B) Using the answer provided by the book for part A
y=Asin(kx - wt + phi)
0.02 m = (0.0215m)sin(0 + 0 + phi)
sinphi = 0.93
phi = 1.20

Once again, this answer is wrong. Book gives phi = 1.95
Yikes!! Any suggestions?

Thank you!!
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Old Jun 27th 2009, 01:56 PM   #2
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Location: Mumbai,India
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The general equation of a wave travelling along negative x-axis


given T of the wave=25 millisecond
i.e. w=2(pi)/T

given for x=0 n t=0

i.e. Asin(phi)=0.02........(1)

Further, particle velocity is specified

so, you can find tan(phi) by dividing (1) by (2)
hence the phase angle can also be found.

For the first, use the value of phi in one of the equations for the answer.

Your mistake is that in the first section, you have assumed the particle at x=0 and t=0 to have a phase 0, i.e. you assumed that phi=0, hence the error.

In the second part the discrepancy is because you got the Principal Solution of the equation.
Since trigonometric functions are periodic, the value of sin1.2 and sin 1.95 are the same .
The appropriate value is decided by the value of the cosine of the phi angle, which is negative [from (2)]
and since the cosine of an obtuse angle is negative, the phi is obtuse and 1.95 and not 1.2

Last edited by Akshay; Jun 27th 2009 at 02:00 PM.
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