Question

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

Answer 1

Step 1. Given data:

Distance between the particles = $2.0cm$

Time taken from mean to crest = $5.0ms$

Step 2. Apply frequency formula and put the values:

Time taken to reach from the mean position to the extreme position,

$\frac{T}{4}=5ms$

Time period of the wave, $$\begin{gathered} T=4\times 5ms \\ T=2\times {10}^{-2}s \end{gathered}$$

Frequency of vibration, $\begin{array}{rcl}f& =& \frac{1}{T}\end{array}$

$$\begin{gathered} f=\frac{1}{2\times {10}^{-2}} \\ f=50Hz \end{gathered}$$

Step 3. Apply wave speed formula and put the values:

Any two neighbouring mean positions always remain at half of the wave length, $\lambda$

$\lambda =2\times$Distance between the particles

$$\begin{gathered} \lambda =2\times 2 \\ \lambda =4cm \end{gathered}$$

Step 4. Apply the wavelength concept and find out the wavelength:

Now, wave speed, $v=\lambda f$

$$\begin{gathered} v=4\times {10}^{-2}m\times 50Hz \\ v=200\times {10}^{-2} \\ v=2m{s}^{-1} \end{gathered}$$

Hence, the frequency, the wavelength and the wave speed are $50Hz,4cm\&2m{s}^{-1}$