A wave of frequency 500 Hz has velocity 360 m/sec. The distance between two nearest points `60^(@)` out of phase, is

Options

A. 70 cm
B. 0.7 m
C. 12.0 cm
D. 120.0 cm

Answer

To solve the problem, we need to find the distance between two nearest points that are 60 degrees out of phase in a wave with a frequency of 500 Hz and a velocity of 360 m/s. Here’s how we can approach this step by step:

### Step 1: Calculate the Wavelength
We know that the velocity of a wave (v) is related to its wavelength (λ) and frequency (f) by the formula:
\[ v = f \cdot \lambda \]

Rearranging this formula to find the wavelength gives us:
\[ \lambda = \frac{v}{f} \]

Substituting the given values:
- \( v = 360 \, \text{m/s} \)
- \( f = 500 \, \text{Hz} \)

We get:
\[ \lambda = \frac{360 \, \text{m/s}}{500 \, \text{Hz}} = \frac{360}{500} = 0.72 \, \text{m} \]

### Step 2: Convert Phase Difference to Radians
The phase difference (Δφ) given is 60 degrees. We need to convert this to radians because the formula we will use requires radians.
\[ \Delta \phi = 60^\circ = \frac{60 \cdot \pi}{180} = \frac{\pi}{3} \, \text{radians} \]

### Step 3: Calculate the Path Difference
The path difference (Δx) corresponding to a phase difference (Δφ) is given by the formula:
\[ \Delta x = \frac{\lambda}{2\pi} \cdot \Delta \phi \]

Substituting the values we have:
\[ \Delta x = \frac{0.72 \, \text{m}}{2\pi} \cdot \frac{\pi}{3} \]

### Step 4: Simplify the Expression
Now, we can simplify this expression:
\[ \Delta x = \frac{0.72}{2 \cdot 3} = \frac{0.72}{6} \]

Calculating this gives:
\[ \Delta x = 0.12 \, \text{m} \]

### Step 5: Convert to Centimeters
Since the problem asks for the distance in centimeters, we convert meters to centimeters:
\[ 0.12 \, \text{m} = 12 \, \text{cm} \]

### Final Answer
Thus, the distance between two nearest points that are 60 degrees out of phase is **12 cm**.