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) = 0.18 sin(2.4x + 17.6t)
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? _________________ 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 = 3.5m and t = 0.43 seconds , what is the velocity of the rope? _______________m/s
At x = 3.5m and t= 0.43 seconds , what is the acceleration of the rope? ______________ m/s^2
What is the average velocity of the rope during one complete oscillation of the rope? ________________ m/s
In what direction is the wave traveling?
On the same rope, how would increasing the wavelength of the wave change the period of oscillation?

Explanation / Answer

1)Amplitude of the wave is 0.18m as maximum value of sin is 1

2)Comparing y(x,t) = 0.18 sin(2.4x + 17.6t) to y = A sin(kx+wt)

where A is amplitude,k=2*(pi)/lamda , w = angular frequency

here w = 2*pi*f = 17.6 => f = 2.8/s

3)k = 2.4 = 2*(pi)/lamda

lamda = 0.38m

4)speed v = f*lamda = 1.064 m/s

5)Tension = Force on rope = m.g.l = mass density*g = 0.124 x 9.8 = 1.215 N

6) velocity of rope = dy/dt = 0.18x17.6xcos(2.4x + 17.6t)

Substituting x = 3.5m and t = 0.43 seconds

v = 3.04 m/s

7) Average value of velocity is zero over one complete oscillation as it is symmetric about x-axis
and we consider above to be +ve and below to be -ve,both cancels out leaving zero.

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