Question

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.

Answer 1

Wave Speed is 2 m/s

Explanation:

Time taken to reach from the mean position to the extreme position, $\frac {T}{4}$= 5ms

Time period (T) of the wave:

T=$4 \times 5\ ms$

= $20 \times 10^-^3$

= $2 \times 10^-^2$ s

Wavelength$(\lambda) = 2 \times Distance\ between\ two\ mean\ positions$

                                       = $2 \times 2$ cm = 4 cm      

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

                     = $\frac {1}{2 \times 10^-^2}$

                     = 50 Hz  

Wave Speed, v = $\lambda f$

= $4 \times 10^-^2 \times 50$

= $200 \times 10^-^2$

= 2 m/s

Answer 2

Time taken to reach from the mean position to the extreme position, \frac {T}{4}= 5ms

Time period (T) of the wave:

T=4 \times 5\ ms

= 20 \times 10^-^3

= 2 \times 10^-^2 s

Wavelength(\lambda) = 2 \times Distance\ between\ two\ mean\ positions

= 2 \times 2 cm = 4 cm

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

= \frac {1}{2 \times 10^-^2}

= 50 Hz

Wave Speed, v = \lambda f

= 4 \times 10^-^2 \times 50

= 200 \times 10^-^2

= 2 m/s

Hope it helps :)