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Question
(i) For the wave on a string described in Exercise 15.11, do all the points on the string oscillate with the same (a) frequency, (b) phase, (c) amplitude? Explain your answers. (ii) What is the amplitude of a point 0.375 m away from one end?
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Solution 1
(i) For the wave on the string described in questions we have seen that l = 1.5 m and λ = 3 m. So, it is clear that λ = λ /2 and for a string clamped at both ends, it is possible only when both ends behave as nodes and there is only one antinode in between i.e., whole string is vibrating in one segment only.
(a) Yes, all the sring particles, except nodes, vibrate with the same frequency v = 60 Hz.
(b) As all string particles lie in one segment, all of them are in same phase.
(c) Amplitude varies from particle to particle. At antinode, amplitude = 2A = 0.06 m. It gradually falls on going towards nodes and at nodes, and at nodes, it is zero.
(ii) Amplitude at a point x = 0.375 m will be obtained by putting cos (120 πt) as + 1 in the wave equation.
`:. A(x) = 0.06 sin((2pi)/3 xx 0.375) xx 1 = 0.06 sin pi/4= 0.042 m`
Solution 2
(i)
(a) Yes, except at the nodes
(b) Yes, except at the nodes
(c) No
(ii) 0.042 m
Explanation:
(i)
(a) All the points on the string oscillate with the same frequency, except at the nodes which have zero frequency.
(b) All the points in any vibrating loop have the same phase, except at the nodes.
(c) All the points in any vibrating loop have different amplitudes of vibration.
(ii) The given equation is:
`y(x,t) = 0.06 sin ((2pi)/3 x) cos(120 pi t)`
For x = 0.375 m and t = 0
Amplitude = Displacement = `0.06 sin((2pi)/3 x)cos 0`
`= 0.06 sin (2pi/3 xx 0.375) xx 1`
`= 0.06 sin (0.25 pi) = 0.06 sin (pi/4)`
`= 0.06 xx 1/sqrt2 = 0.042 m`
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