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Transverse waves are traveling on a long string that is under a tension of 10.0 N. The equation describing these waves is y(x,t)=( 6.05 cm)sin[( 435 s−1)t−( 45.1 m−1)x]. Find the linear mass density of this string.

Answer :

Answer:

μ = 0.108 kg/m

Explanation:

Given that

Tension ,T= 10 N

y(x,t)=( 6.05 cm)sin[( 435 s−1)t−( 45.1 m−1)x]

Standard form

Y(x,t)=A sin[ωt−k x]

By compare ,we can say that

ω = 435 s⁻¹

We know that   ω = 2 π f

[tex]f=\dfrac{\omega}{2\pi }[/tex]

[tex]f=\dfrac{435}{2\pi }\ Hz[/tex]

f=69.23  Hz

k = 45.1 m⁻¹

We know that

[tex]\lambda =\dfrac{2 \pi}{k}[/tex]

[tex]\lambda =\dfrac{2 \pi}{45.1}[/tex]

λ =0.139  m

We know that the velocity given as

V = f λ

V= 0.139 x 69.23 m/s

V= 9.6 m/s

The linear mass density given as

[tex]\mu=\dfrac{T}{V^2}\ kg/m[/tex]

[tex]\mu=\dfrac{10}{9.6^2}\ kg/m[/tex]

μ = 0.108 kg/m

This is the mass density.

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