The equation of the standing wave in a string clamped at both ends, vibrating in its third harmonic is given by
`y=0.4sin(0.314x)cos(600pit)`
where, `x` and `y` are in cm and `t` in sec.
`(a)` the frequency of vibration is `300Hz`
`(b)` the length of the string is `30cm`
`(c )` the nodes are located at `x=0`, `10cm`, `30cm`

Options

A. Only a is true
B. `a,b` are true
C. `b,c` are true
D. `a,b,c` are true

Explanation

To solve the problem, we will analyze the given equation of the standing wave and extract the required information step by step.

### Given:
The equation of the standing wave is:
\[ y = 0.4 \sin(0.314x) \cos(600\pi t) \]
where \( x \) and \( y \) are in cm and \( t \) is in seconds.

### Step 1: Determine the frequency of vibration
The angular frequency \( \omega \) is given by the coefficient of \( t \) in the cosine term:
\[ \omega = 600\pi \]
The linear frequency \( f \) can be found using the relationship:
\[ \omega = 2\pi f \]
Thus,
\[ 600\pi = 2\pi f \]
Dividing both sides by \( 2\pi \):
\[ f = \frac{600\pi}{2\pi} = 300 \text{ Hz} \]

### Step 2: Determine the length of the string
The wave number \( k \) is given by the coefficient of \( x \) in the sine term:
\[ k = 0.314 \]
The wavelength \( \lambda \) can be calculated using the relationship:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} \]
Substituting \( k \):
\[ \lambda = \frac{2\pi}{0.314} \approx 20 \text{ cm} \]
In a string clamped at both ends, the length \( L \) of the string in the nth harmonic is given by:
\[ L = \frac{n\lambda}{2} \]
For the third harmonic (\( n = 3 \)):
\[ L = \frac{3 \cdot 20}{2} = 30 \text{ cm} \]

### Step 3: Determine the positions of the nodes
Nodes occur at positions where the sine function is zero:
\[ \sin(0.314x) = 0 \]
This occurs when:
\[ 0.314x = n\pi \]
where \( n \) is an integer. Solving for \( x \):
\[ x = \frac{n\pi}{0.314} \]
Calculating for \( n = 0, 1, 2, \ldots \):
- For \( n = 0 \): \( x = 0 \)
- For \( n = 1 \): \( x = \frac{\pi}{0.314} \approx 10 \text{ cm} \)
- For \( n = 2 \): \( x = \frac{2\pi}{0.314} \approx 20 \text{ cm} \)
- For \( n = 3 \): \( x = \frac{3\pi}{0.314} \approx 30 \text{ cm} \)

Thus, the nodes are located at \( x = 0 \), \( x = 10 \text{ cm} \), and \( x = 30 \text{ cm} \).

### Summary of Results:
(a) The frequency of vibration is \( 300 \text{ Hz} \).
(b) The length of the string is \( 30 \text{ cm} \).
(c) The nodes are located at \( x = 0 \), \( x = 10 \text{ cm} \), and \( x = 30 \text{ cm} \).

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