A horizontal wire is under a tension of 345 N and has a mass per unit length of 6.85 10-3 kg/m. A transverse wave with an amplitude of 2.65 mm and a frequency of 525 Hz is traveling on this wire. As the wave passes, a particle of the wire moves up and down in simple harmonic motion.
(a) Obtain the speed of the wave. (m/s)
(b) Obtain the maximum speed with which the particle moves up and down.(m/s)
(a) Obtain the speed of the wave. (m/s)
(b) Obtain the maximum speed with which the particle moves up and down.(m/s)
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(a) the speed of the wave
Use The Melde's law:
v = √F/μ = √345/6.85 10^-3 = 224.42 m/s
(b) the maximum speed with which the particle moves up and down:
vy = ωA = 2πf * A =2π* 525 * 2.65 x 10^-3 = 2.78π m/s
Use The Melde's law:
v = √F/μ = √345/6.85 10^-3 = 224.42 m/s
(b) the maximum speed with which the particle moves up and down:
vy = ωA = 2πf * A =2π* 525 * 2.65 x 10^-3 = 2.78π m/s