A periodic transverse wave is established on a string such that there are exactly two cycles traveling along a 3.0-m section of the string. The crests move at 20.0 m/s.
How long does it take a particle at the top of a crest to reach the bottom of an adjacent trough?

Options

A. 0.018s
B. 0.038s
C. 0.075s
D. 0.150s

Answer

To solve the problem, we need to determine how long it takes for a particle at the top of a crest to reach the bottom of an adjacent trough in a periodic transverse wave.

### Step-by-Step Solution:

1. **Identify the Wavelength**:
- The problem states that there are 2 cycles (or wavelengths) in a 3.0 m section of the string.
- Therefore, the wavelength (λ) can be calculated as:
\[ \text{Wavelength} (\lambda) = \frac{\text{Total Length}}{\text{Number of Cycles}} = \frac{3.0 \, \text{m}}{2} = 1.5 \, \text{m} \]

2. **Determine the Wave Speed**:
- The speed (v) of the wave is given as 20.0 m/s.

3. **Calculate the Frequency**:
- The frequency (f) of the wave can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
- Substituting the known values:
\[ f = \frac{20.0 \, \text{m/s}}{1.5 \, \text{m}} \approx 13.33 \, \text{Hz} \]

4. **Calculate the Time Period**:
- The time period (T) is the reciprocal of the frequency:
\[ T = \frac{1}{f} = \frac{1}{13.33 \, \text{Hz}} \approx 0.075 \, \text{s} \]

5. **Determine the Time to Move from Crest to Trough**:
- The time taken for a particle to move from the top of a crest to the bottom of an adjacent trough is half of the time period (T/2):
\[ t = \frac{T}{2} = \frac{0.075 \, \text{s}}{2} \approx 0.0375 \, \text{s} \]

### Final Answer:
The time it takes for a particle at the top of a crest to reach the bottom of an adjacent trough is approximately **0.0375 seconds**.

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