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Q14.11: NCERT Physics - Part 2 WAVES 14.11 Solution: The transverse displacement of a string (clamped at its both…

The transverse displacement of a string (clamped at its both ends) is given by y(x, t) = 0.06 sin(2�x/3) cos(120�t) where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 _ 10__ kg. Answer the following: (a) Does the function represent a travelling wave or a stationary wave? (b) Interpret the wave as a superposition of two waves travelling in opposite directions. What is the wavelength, frequency, and speed of each wave? (c) Determine the tension in the string.

Step-by-Step Verified Solution

**Answer:** (a) Stationary wave. (b) λ=3\lambda = 3 m, f=60f = 60 Hz, v=180v = 180 m/s. (c) T=648T = 648 N. **Step-by-Step Explanation:** (a) The equation has the form y=Asin⁡(kx)cos⁡(ωt)y = A\sin(kx)\cos(\omega t), which represents a stationary wave . (b) Comparing with y=2Asin⁡(kx)cos⁡(ωt)y = 2A\sin(kx)\cos(\omega t): k=2π/3k = 2\pi/3, so λ=3\lambda = 3 m. ω=120π\omega = 120\pi, so f=60f = 60 Hz. Speed v=fλ=180v = f\lambda = 180 m/s. The two component waves have the same wavelength, frequency, and speed, but travel in opposite directions [citation:2]. (c) Linear mass density μ=m/l=0.03/1.5=0.02\mu = m/l = 0.03/1.5 = 0.02 kg/m. Tension T=μv2=0.02×(180)2=648T = \mu v^2 = 0.02 \times (180)^2 = 648 N.

All Questions in this Chapter

Q14.1 Q14.2 Q14.3 Q14.4 Q14.5 Q14.6 Q14.7 Q14.8 Q14.9 Q14.10 Q14.11 Q14.12 Q14.13 Q14.14 Q14.15 Q14.16 Q14.17 Q14.18 Q14.19

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