A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.

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Solution

Time period, $T = 4 \times 5 m s$

$= 20 \times 10^{- 3} = 2 \times 10^{- 2} s$

$λ = 2 \times 2 c m = 4 c m$

Frequency, $f = \frac{1}{T}$

$= \frac{1}{(2 \times 10^{- 2})} = 50 s^{- 1}$

$= 50 H z$

Wave speed, $λ f = 4 \times 10^{- 2} \times 50 m / s$

$= 200 \times 10^{- 2} m / s = 2 m / s$

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A particle on a stretched string supporting a travelling wave, takes 5.0 ms to move from its mean position to the extreme position. The distance between two consecutive particles, which are at their mean positions, is 2.0 cm. Find the frequency, the wavelength and the wave speed.

Time period, T=4×5 ms =20×10−3=2×10−2s λ=2×2 cm=4 cm Frequency, f=1T =1(2×10−2)=50s−1 =50 Hz Wave speed, λf=4×10−2×50 m/s =200×10−2m/s=2 m/s

Time period, $T = 4 \times 5 m s$

$= 20 \times 10^{- 3} = 2 \times 10^{- 2} s$

$λ = 2 \times 2 c m = 4 c m$

Frequency, $f = \frac{1}{T}$

$= \frac{1}{(2 \times 10^{- 2})} = 50 s^{- 1}$

$= 50 H z$

Wave speed, $λ f = 4 \times 10^{- 2} \times 50 m / s$

$= 200 \times 10^{- 2} m / s = 2 m / s$