Verified Solution • 2025–26 Session moderate

Q14.8: NCERT Physics - Part 2 WAVES 14.8 Solution: A transverse harmonic wave on a string is described by y(x, t)…

A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36 t + 0.018 x + �/4) where x and y are in cm and t in s. The positive direction of x is from left to right. (a) Is this a travelling wave or a stationary wave? If it is travelling, what are the speed and direction of its propagation? (b) What are its amplitude and frequency? (c) What is the initial phase at the origin? (d) What is the least distance between two successive crests in the wave?

Step-by-Step Verified Solution

**Answer:** (a) Travelling wave, v=20v = 20 m/s, right to left. (b) A=3.0A = 3.0 cm, f=5.73f = 5.73 Hz. (c) ϕ=π/4\phi = \pi/4. (d) d=3.49d = 3.49 m. **Step-by-Step Explanation:** (a) The equation has the form y=Asin⁡(ωt+kx+ϕ)y = A\sin(\omega t + kx + \phi) with the same sign for ωt\omega t and kxkx, indicating propagation in the negative x-direction (right to left). Speed v=ω/k=36/0.018=2000v = \omega/k = 36/0.018 = 2000 cm/s =20= 20 m/s . (b) Amplitude A=3.0A = 3.0 cm. Frequency f=ω/(2π)=36/(2π)=5.73f = \omega/(2\pi) = 36/(2\pi) = 5.73 Hz. (c) At x=0,t=0x=0, t=0, phase is ϕ=π/4\phi = \pi/4. (d) Wavelength λ=2π/k=2π/0.018=349\lambda = 2\pi/k = 2\pi/0.018 = 349 cm =3.49= 3.49 m. The least distance between successive crests is λ=3.49\lambda = 3.49 m [citation:5].

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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