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 find the frequency, wavelength, and wave speed of the wave on the stretched string.

### Step 1: Determine the Time Period of the Wave

We know that the time taken for a particle to move from its mean position to its extreme position is given as 5.0 ms. This movement represents a quarter of a complete cycle (from mean to extreme position).

1. **Given**: Time from mean to extreme position = 5.0 ms = 5 × 10^(-3) s
2. **Since this is 1/4 of the time period (T)**:
\[ \frac{T}{4} = 5 \times 10^{-3} \text{ s} \]
Therefore, the time period \( T \) is:
\[ T = 4 \times 5 \times 10^{-3} = 20 \times 10^{-3} \text{ s} = 2 \times 10^{-2} \text{ s} \]

### Step 2: Calculate the Frequency

Frequency \( f \) is the reciprocal of the time period \( T \):
\[ f = \frac{1}{T} \]
Substituting the value of \( T \):
\[ f = \frac{1}{2 \times 10^{-2}} = 50 \text{ Hz} \]

### Step 3: Determine the Wavelength

The distance between two consecutive particles at their mean positions is given as 2.0 cm. This distance corresponds to half of the wavelength \( \lambda \) (since the mean positions are at the midpoint of the wave).

1. **Given**: Distance between mean positions = 2.0 cm
2. Therefore:
\[ \frac{\lambda}{2} = 2.0 \text{ cm} \]
Thus, the wavelength \( \lambda \) is:
\[ \lambda = 2 \times 2.0 \text{ cm} = 4.0 \text{ cm} \]

### Step 4: Calculate the Wave Speed

The wave speed \( v \) can be calculated using the formula:
\[ v = f \times \lambda \]
Substituting the values of \( f \) and \( \lambda \):
1. Convert \( \lambda \) to meters for consistency:
\[ \lambda = 4.0 \text{ cm} = 0.04 \text{ m} \]
2. Now calculate the wave speed:
\[ v = 50 \text{ Hz} \times 0.04 \text{ m} = 2.0 \text{ m/s} \]

### Final Results
- Frequency \( f = 50 \text{ Hz} \)
- Wavelength \( \lambda = 4.0 \text{ cm} \)
- Wave Speed \( v = 2.0 \text{ m/s} \)

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