Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 343 m/s, piano A produces a wavelength of 0.769 m, while piano B produces a wavelength of 0.776 m. How much time separates successive beats?

Options

A. 1.4 s
B. 0.76 s
C. 0.25 s
D. 0.07 s

Answer

To solve the problem of determining how much time separates successive beats produced by two out-of-tune pianos, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Speed of sound (v) = 343 m/s
- Wavelength of Piano A (λ₁) = 0.769 m
- Wavelength of Piano B (λ₂) = 0.776 m

2. **Calculate the Frequencies of the Sounds**:
- The frequency (f) of a wave can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
- For Piano A:
\[ f_1 = \frac{343 \, \text{m/s}}{0.769 \, \text{m}} \approx 446 \, \text{Hz} \]
- For Piano B:
\[ f_2 = \frac{343 \, \text{m/s}}{0.776 \, \text{m}} \approx 442 \, \text{Hz} \]

3. **Calculate the Beat Frequency**:
- The beat frequency (f_beat) is the absolute difference between the two frequencies:
\[ f_{\text{beat}} = |f_1 - f_2| = |446 \, \text{Hz} - 442 \, \text{Hz}| = 4 \, \text{Hz} \]

4. **Calculate the Time Period of the Beat Frequency**:
- The time period (T) of the beat frequency is the reciprocal of the beat frequency:
\[ T = \frac{1}{f_{\text{beat}}} = \frac{1}{4 \, \text{Hz}} = 0.25 \, \text{seconds} \]

### Final Answer:
The time that separates successive beats is **0.25 seconds**.

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