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.

Explanation

To solve the problem step by step, we will follow the information given in the question and the video transcript.

### Step 1: Determine the time period of the wave

The time taken for a particle to move from its mean position to its extreme position is given as 5.0 ms. This represents one-fourth of the time period (T) of the wave, as it takes 5 ms to go from the mean position to the extreme position, and then another 5 ms to return to the mean position, and so on.

\[ \text{Time period (T)} = 4 \times \text{time from mean to extreme} = 4 \times 5 \, \text{ms} = 20 \, \text{ms} = 20 \times 10^{-3} \, \text{s} \]

### Step 2: Calculate the frequency of the wave

Frequency (f) is the reciprocal of the time period (T).

\[ f = \frac{1}{T} = \frac{1}{20 \times 10^{-3}} = \frac{1}{0.020} = 50 \, \text{Hz} \]

### Step 3: Determine the wavelength of the wave

The distance between two consecutive particles at their mean positions is given as 2.0 cm. This distance represents half of the wavelength (λ) because it is the distance between two points that are in phase (both at mean position).

\[ \text{Wavelength (λ)} = 2 \times \text{distance between mean positions} = 2 \times 2.0 \, \text{cm} = 4.0 \, \text{cm} = 4.0 \times 10^{-2} \, \text{m} \]

### Step 4: Calculate the wave speed

Wave speed (v) can be calculated using the formula:

\[ v = f \times \lambda \]

Substituting the values we found:

\[ v = 50 \, \text{Hz} \times 4.0 \times 10^{-2} \, \text{m} = 2.0 \, \text{m/s} \]

### Summary of Results

- Frequency (f) = 50 Hz
- Wavelength (λ) = 4.0 cm (or 0.04 m)
- Wave Speed (v) = 2.0 m/s