Vibration from a 525 Hz tuning fork sets up standing waves in a string clamped a

Question

Vibration from a 525 Hz tuning fork sets up standing waves in a string clamped at both ends. The wave speed for the string is 470m/s. The standing wave has four loops and an amplitude of 2.0mm.

(a) What is the length of the string?

(b) What is the equation for the displacement of the string as a function of position and time?

a. y(x,t) = 0.002sin(7.0x)cos(3300t)

b. y(x,t) = 0.004sin(14.0x)cos(3300t)

c. none of these

d. y(x,t) = 0.002sin(3.5x)cos(6600t)

I'm not sure if I got (a) or not. I found out the wavelength first by using the speed v (470m/s) and the frequency f (525Hz) using the equation: [Wavelength = speed / frequency] and I got the value 0.895m. From there I used the equation [L = (n/2) * wavelength] where n = 4 loops. Using that I got the length of the string as 1.79m. Please check my work on that one.

For (b) I really don't know where to start.

Thanks.

Explanation / Answer

a) You have got a) right.

b)amplitude=0.002m. Therefore, it is either a) or d).

The standing wave equation is y=2y0Cos(wt)Sin(kx)

w=2pi*f=2pi*525=33oo rad/s

Therefore, the cos term becomes cos(3300t)

Also the sin term is Sin (kx)

k=2pi/lamda=7/m

Therefore, a) is correct

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