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Standard XII

Physics

Velocity Displacement Relationship

A long string...

Question

A long string having a cross-sectional area $0.80 m m^{2}$ and density, $12.5 g / c c$ is subjected to a tension of $64 N$ along the positive x-axis. One end of this string is attached to a vibrator at $x = 0$ moving in transverse direction at a frequency of $20 H z$ . At $t = 0$ , the source is at a maximum displacement $y = 1.0 c m .$ Find the speed of the wave travelling on the string.

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Solution

Mass per unit length of the string= $μ = ρ A = 0.8 \times 10^{- 6} \times 12.5 \times 10^{3} = 0.01 k g / m$
Thus speed of the wave= $\sqrt{\frac{T}{μ}} = \sqrt{\frac{64}{0.01}} = 80 m / s$

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Mass per unit length of the string=μ=ρA=0.8×10−6×12.5×103=0.01kg/mThus speed of the wave=√Tμ=√640.01=80m/s

Mass per unit length of the string= $μ = ρ A = 0.8 \times 10^{- 6} \times 12.5 \times 10^{3} = 0.01 k g / m$
Thus speed of the wave= $\sqrt{\frac{T}{μ}} = \sqrt{\frac{64}{0.01}} = 80 m / s$