How Do You Calculate Transverse Speed and Acceleration for a Wave on a String?

[complexc25]

A transverse wave on a string is described by the following wave function.
y = (0.100 m) sin [(x/11 + 3t)]

• (a) Determine the transverse speed and acceleration at t = 0.150 s for the point on the string located at x = 1.60 m.
  
• (b) What are the wavelength, period, and speed of propagation of this wave?
• wrong check mark

I know that to find the transverse speed i need to find the derivative. I need a refresher on the derivative of that sin function, because I am getting the wrong answer. Once i get part A ill find out part B.
thanks in advance

 

[Kurdt]

The derivative of sin(x) is cos(x).

 

[Moetasim]

Hello,

I can help you with this problem. Let's start with part A. To find the transverse speed, we need to take the derivative of the wave function with respect to time, which is given as:

y' = (0.100 m)(3)cos[(x/11 + 3t)]

Plugging in the given values, we get:

y'(1.60, 0.150) = (0.100 m)(3)cos[(1.60/11 + 3(0.150))] = 0.026 m/s

This is the transverse speed at t = 0.150 s for the point on the string located at x = 1.60 m.

To find the transverse acceleration, we need to take the second derivative of the wave function with respect to time, which is given as:

y'' = -(0.100 m)(3)^2sin[(x/11 + 3t)]

Plugging in the given values, we get:

y''(1.60, 0.150) = -(0.100 m)(3)^2sin[(1.60/11 + 3(0.150))] = -0.1 m/s^2

This is the transverse acceleration at t = 0.150 s for the point on the string located at x = 1.60 m.

Moving on to part B, the wavelength is given by the distance between two consecutive points on the string that have the same displacement and are in phase. In this case, it is the distance between two consecutive peaks or troughs of the wave. From the given wave function, we can see that the wavelength is 11 m.

The period is the time taken for one complete cycle of the wave. In this case, it is the time taken for the wave to travel one wavelength. From the wave function, we can see that the period is 3 seconds.

The speed of propagation is the speed at which the wave travels along the string. It is given by the product of wavelength and frequency. From the given wave function, we can see that the frequency is 1/3 Hz, so the speed of propagation is (11 m)(1/3 Hz) = 3.67 m/s.

I hope this helps. Let me know if you have any further questions. Good luck with your calculations!