A vibrating string 150 cm long is under a tension of 1 N. The results from five
ID: 249439 • Letter: A
Question
A vibrating string 150 cm long is under a tension of 1 N. The results from five consecutive stroboscopic pictures are shown in the figure. the pictures were taken at a rate of 5000 flashes per min. the maximum displacement appears at picture 1 and 5 and no other maxima between them. Find the period, frequency and wave length of the travelling on this string. In which harmonics the string is vibrating? What is the speed of the travelling wave? What is the speed of the point P at position 1? What is the speed of the point P at position 3?Explanation / Answer
(a) Here, wavelength is equal to length of string, = 150/100 m = 1.5 m
Four flashes occur from position 1 to position 5
So, Time Period ,T is given by :-
T/2 = 4 * 60/5000 sec
T/2 =0.048
So, T = 0.096 sec
Frequency,f = 1/T
=1/0.096
= 10.41 Hz
So, Wavelength of wave = 1.5 m
Time period of wave = 0.096 sec
Frequency of wave = 10.41 Hz
(b) String is vibrating in 2nd Harmonic or 1st overtone. The fundamental standing wave has nodes at each end and no nodes in between. This standing wave has one additional node.
(c) speed of travelling wave = frequency * wavelength
=10.41 *1.5 m/sec
=15.615 m/sec
(d) In position 1, point P is at its maximum displacement and its speed is zero .
(e) In position 3 , point P is passing through its equilibrium position .So, it has maximum velocity,
Vmax = 2*3.14 *frequency*amplitude
= 2*3.14*10.41*0.12
=7.84 m/sec
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