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A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/s. Take the origin at the left end of the string has its maximum upward displacement.

(a) What are the frequency, angular frequency, and wave number of the wave?

(b) What is the function y(x,t) that describes the wave?

(c) What is y(t)for a particle at the left end of the string?

(d) What is y(t) for a particular 1.35 m to right of the origin?

(e) What is the maximum magnitude of transverse velocity of any particle of the string?

(f) Find the transverse displacement and the transverse velocity of a particle 1.35 m to the right of the origin at time t=0.0625s.

Short Answer

Expert verified

Thus, (a) the frequency is 20Hz, angular frequency is 40ฯ€sโˆ’1 and wave number is 3.49mโˆ’1.

Step by step solution

01

(a) Given in the question

Amplitude is A=2.50mm

Wavelength is ฮป=1.80m.

Speed of propagation is v=36.0m/s.

02

Standard wave equation

The standard wave equation is:

y(x,t)=Acosโก(kx+ฯ‰t)

Here, A is amplitude, ฯ‰ is angular frequency, k is wave number.

Formula for period is:

ฯ‰=2ฯ€f

Formula of frequency is:

f=1T

Formula of wave number is:

k=2ฯ€ฮป

Formula of speed of propagation:

v=fฮป

03

Calculate frequency, angular frequency, and wave number

Use the formula to calculate frequency,

f=vฮป=36m/s1.80m=20sโˆ’1=20Hz

Use the formula to calculate angular frequency,

ฯ‰=2ฯ€f=2ฯ€ร—20=40ฯ€sโˆ’1

Use the formula to calculate wave number,

ฮป=2ฯ€k1.80m=2ฯ€kk=2ฯ€1.80m=3.49mโˆ’1

Hence, the frequency is 20Hz, angular frequency is 40ฯ€sโˆ’1 and wave number is 3.49mโˆ’1.

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