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.
What is the shortest horizontal distance from a crest to a point of zero acceleration?

Options

A. 0.38m
B. 0.75m
C. 1.5m
D. 3.0m

Answer

To solve the problem step by step, we can follow these instructions:

### Step 1: Determine the wavelength (λ)
Given that there are 2 cycles in a 3.0 m section of the string, we can find the wavelength by dividing the total distance by the number of cycles.

\[ \text{Wavelength} (\lambda) = \frac{\text{Total distance}}{\text{Number of cycles}} = \frac{3.0 \, \text{m}}{2} = 1.5 \, \text{m} \]

### Step 2: Understand the wave motion
The wave is a transverse wave, and the particles of the string undergo simple harmonic motion (SHM). The acceleration of a particle in SHM is given by the formula:

\[ a = -\omega^2 x \]

where \( x \) is the displacement from the mean position. The acceleration is zero when \( x = 0 \), which corresponds to the mean position of the wave.

### Step 3: Identify the distance from the crest to the mean position
In a sine wave, the distance from a crest to the mean position is one-quarter of the wavelength. Therefore, we can express this distance as:

\[ \text{Distance from crest to mean position} = \frac{\lambda}{4} \]

### Step 4: Calculate the distance
Now that we know the wavelength \( \lambda = 1.5 \, \text{m} \), we can calculate the distance from the crest to the point of zero acceleration:

\[ D = \frac{\lambda}{4} = \frac{1.5 \, \text{m}}{4} = 0.375 \, \text{m} \]

### Step 5: Round the answer
Rounding the answer to two decimal places, we get:

\[ D \approx 0.38 \, \text{m} \]

### Final Answer
The shortest horizontal distance from a crest to a point of zero acceleration is approximately **0.38 m**.

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