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Worked Examples · Example 6.1

Q.A string 105 cm long is fixed at one end. Transverse vibrations of frequency 15 Hz are imposed at the free end. A stationary wave, produced in the string, consists of 3 loops. Calculate the speed of progressive waves which have produced the stationary wave in the string.

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✓ Free question

l = 3λ/2 → λ = (2/3)×105 = 70 cm = 0.70 m; v = 15×0.70 = 10.50 m/s.

With 3 loops the length holds three half-wavelengths:

l=3λ2⇒λ=23l=23×105=70 cm=0.70 ml = 3\frac{\lambda}{2} \Rightarrow \lambda = \frac{2}{3}l = \frac{2}{3}\times 105 = 70\ \mathrm{cm} = 0.70\ \mathrm{m}

v=nλ=15×0.70=10.50 m s−1v = n\lambda = 15 \times 0.70 = 10.50\ \mathrm{m\,s^{-1}}

✓Final answer

λ = 0.70 m, v = nλ = 15 × 0.70 = 10.50 m s⁻¹.

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