Question

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

y(x,t) = 0.12sin(2.1x + 17.5t)

The mass density of the rope is μ = 0.114 kg/m. x and y are measured in meters and t in seconds.

What is the amplitude of the wave? ___m

What is the frequency of oscillation of the wave?____ Hz

What is the wavelength of the wave?___m

What is the speed of the wave? _____m/s

What is the tension in the rope? ____N

At x = 3.8 m and t = 0.44 s, what is the velocity of the rope? (watch your sign)___m/s

At x = 3.8 m and t = 0.44 s, what is the acceleration of the rope? (watch your sign) _____m/s^2

What is the average speed of the rope during one complete oscillation of the rope? ______m/s

PLEASE HELP

Answer 1

y(x,t) = 0.12sin(2.1x + 17.5t)

this is of the form,

whose Amplitude is A, wavevector is k and angular frequency is .

So, from the given wave, the amplitude is : A = 0.12m.

Frequency is

Wavelength is:

Speed of the wave will be:

The velocity of the wave exhibiting this sinusoidal motion is given by:

where T is the tension in the rope.

therefore .

Now, consider this displacement function:

differentiating this with time, we get:

[using chain rule]

and dy/dt is nothing but the velocity of the wave.

so,

and so acceleration will be:

therefore our displacement equation: y(x,t) = 0.12sin(2.1x + 17.5t), the velocity function will be:

=> m/s

and

So, at x = 3.8m and t = 0.44s:

v(x,t) = 2.1cos[(2.1 x 3.8) + (17.5 x 0.44)]

=> v(x,t) = 2.1cos[15.68 rad] = - 2.099 m/s

and at x = 3.8 m and t = 0.44s:

= - 36.75sin[15.68 rad] = - 1.0275 m/s².

The average speed of the rope will be: v = 4A/T = 4Af = 4[0.12][2.785] = 1.3368 m/s.