A transverse harmonic wave travels on a rope according to the following expressi

Question

A transverse harmonic wave travels on a rope according to the following expression: y(x,t) = 0.15sin(2.1x + 17.3t) The mass density of the rope is = 0.124 kg/m. x and y are measured in meters and t in seconds. 1) What is the amplitude of the wave? 0.15 m Submit 2) What is the frequency of oscillation of the wave? 2.75 Hz Submit 3) What is the wavelength of the wave? 2.99 m Submit 4) What is the speed of the wave? 8.24 m/s Submit 5) What is the tension in the rope? N Submit 6) At x = 3.2 m and t = 0.4 s, what is the velocity of the rope? (watch your sign) m/s Submit 7) At x = 3.2 m and t = 0.4 s, what is the acceleration of the rope? (watch your sign) 0.08 m/s2 Submit 8) What is the average speed of the rope during one complete oscillation of the rope?

Explanation / Answer

Here ,

y(x,t) = 0.15sin(2.1x + 17.3t)

y = A * sin(k * x + w * t)

1)

comparing , A = 0.15 m

hence, the amplitude of wave is 0.15 m

2)

as w = 17.3 rad/s

frequency of oscillation = w/(2pi)

frequency of oscillation = 17.3/6.282

frequency of oscillation = 2.75 Hz

3)

as k = 2.1 m^-1

wavelength = 2pi/k

wavelength = 2pi/2.1

wavelength = 2.99 m

4)

speed of wave = frquency * wavelength

speed of wave = 2.99 * 2.75 m/s

speed of wave = 8.23 m/s

the speed of wave is 8.23 m/s

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