A string of length `1` is stretched along the `x`-axis and is rigidly clamped at `x=0` and `x=1`. Transverse vibrations are produced in the string. For `n^(th)` harmonic which of the following relations may represents the shape of the string at any time
`(a)` `y=2Acosomegatcos((npix)/(l))`
`(b)` `y=2Asinomegatcos((npix)/(l))`
`(c )``y=2Acosomegatsin((npix)/(l))`
`(d)` `y=2Asinomegatsin((npix)/(l))`

Options

A. `c` only
B. `c` and `d` only
C. `a` only
D. `a,b,c` and `d`

Explanation

To solve the problem, we need to analyze the transverse vibrations of a string that is clamped at both ends. The general form of the equation representing the shape of the string for the nth harmonic can be derived from the boundary conditions and the wave equation.

### Step 1: Understand the Boundary Conditions
The string is clamped at \( x = 0 \) and \( x = 1 \). This means that at these points, the displacement \( y \) must be zero:
- \( y(0, t) = 0 \)
- \( y(1, t) = 0 \)

### Step 2: Write the General Form of the Solution
For a string fixed at both ends, the general solution for transverse vibrations can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t) + B \sin(kx) \sin(\omega t) \]
where:
- \( A \) and \( B \) are constants,
- \( k \) is the wave number,
- \( \omega \) is the angular frequency.

### Step 3: Determine the Wave Number \( k \)
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \]
For the nth harmonic, the wavelength is given by:
\[ \lambda = \frac{2l}{n} \]
where \( l \) is the length of the string. Since \( l = 1 \):
\[ \lambda = \frac{2}{n} \]
Thus, the wave number becomes:
\[ k = \frac{2\pi}{\lambda} = \frac{2\pi n}{2} = n\pi \]

### Step 4: Substitute \( k \) into the Equation
Now substituting \( k \) into the general form, we get:
\[ y(x, t) = A \sin(n\pi x) \cos(\omega t) + B \sin(n\pi x) \sin(\omega t) \]

### Step 5: Simplify the Equation
We can combine the terms to write it in a more compact form. If we assume \( A = 2A \) and \( B = 0 \) (for simplicity), we can express the equation as:
\[ y(x, t) = 2A \sin(n\pi x) \cos(\omega t) \]
This indicates that the shape of the string can be represented as:
\[ y(x, t) = 2A \sin(n\pi x) \cos(\omega t) \]

### Step 6: Identify the Correct Option
From the options provided:
- (a) \( y = 2A \cos(\omega t) \cos\left(\frac{n\pi x}{l}\right) \)
- (b) \( y = 2A \sin(\omega t) \cos\left(\frac{n\pi x}{l}\right) \)
- (c) \( y = 2A \cos(\omega t) \sin\left(\frac{n\pi x}{l}\right) \)
- (d) \( y = 2A \sin(\omega t) \sin\left(\frac{n\pi x}{l}\right) \)

The correct form that matches our derived equation is:
\[ y = 2A \sin(n\pi x) \cos(\omega t) \]
This corresponds to option (c).

### Final Answer
The correct relation that represents the shape of the string at any time for the nth harmonic is:
**(c) \( y = 2A \cos(\omega t) \sin\left(\frac{n\pi x}{l}\right) \)**