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A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an amplitude of 0.157 mm and a frequency of 92.2 Hz and is traveling in the negative direction of an x axis. If the wave equation is of the form y(x,t) = ym sin(kx + ωt), what are (a) ym, (b) k, and (c) ω, and (d) the correct choice of sign in front of ω?

Respuesta :

Answer:

Part a)

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

Explanation:

As we know that the speed of wave in string is given by

[tex]v = \sqrt{\frac{T}{m/L}}[/tex]

so we have

[tex]T = 17.5 N[/tex]

[tex]m/L = 5.4 g/cm = 0.54 kg/m[/tex]

now we have

[tex]v = \sqrt{\frac{17.5}{0.54}}[/tex]

[tex]v = 5.69 m/s[/tex]

now we have

Part a)

[tex]y_m [/tex] = amplitude of wave

[tex]y_m = 0.157 mm[/tex]

part b)

[tex]k = \frac{\omega}{v}[/tex]

here we know that

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(92.2) = 579.3 rad/s[/tex]

so we  have

[tex]k = \frac{579.3}{5.69}[/tex]

[tex]k = 101.8 rad/m[/tex]

Part c)

[tex]\omega = 579.3 rad/s[/tex]

Part d)

here since wave is moving in negative direction so the sign of [tex]\omega[/tex] must be positive

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