Q4. A vibrating string, 0.500m long and fixed at both ends, is under a tension o
ID: 3308740 • Letter: Q
Question
Q4. A vibrating string, 0.500m long and fixed at both ends, is under a tension of 3.00N. A stroboscope was used to take the 5 successive pictures of the vibrating string (shown below). Observations reveal that the maximum displacement (1.50cm) occurred at flashes 1 and 5 with no other maxima between them. The time between successive flashes is 0.0120s (16 pts. total) a. Find the frequency, f, of the vibrating string wave. (2.0 pts.) 1.5 cm 1.5 cm b. Find the wavelength, , for the traveling waves on this string. (2.0 pts.)Explanation / Answer
a) As we can see, this is a standing wave because the aplitudes of each point on the string is fixed. Only two points on the string can achieve a displacement of 1.5cm. The points are called anti-nodes. The two ends of the string and the center of the string do not oscillate at all. These points are called nodes. Also this is the second harmonic of the string where wavelength=length of string=0.5m. Now the time period of the standing wave is the time it would take the string to go from the state in picture 1 to the state in picture 5 and then back to 1. so the time period is T=2x5x0.0120s=0.12s. The frequency is the inverse of time period, so frequency f = 1/T = 1/0.12 = 8.33 Hz
b) The travelling waves are reflected from the boundary and it interferes with the original wave to give standing waves. The two travelling waves have to be the same wavelength so that they can interfere to form a standing wave of that same wavelenth. Therefore the wavelength of the travelling wave is also 0.5m
The speed of the travelling wave can be found as wavelength x frequency and this velocity and the tension of the string can be used to find the mass per unit length of the string
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