A 660 Hz tuning fork sets up vibration in a string clamped at both ends. The wave speed for a transverse wave on this string is `220 m s^-1` and the string vibrates in three loops. (a) Find the length of the string. (b) If the maximum amplitude of a particle is 0.5 cm, write a suitable equation describing the motion.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as stated in the question.

### Part (a): Find the length of the string.

1. **Identify the given values**:
- Frequency (f) = 660 Hz
- Wave speed (v) = 220 m/s
- Number of loops (n) = 3

2. **Use the wave speed formula**:
The relationship between wave speed (v), frequency (f), and wavelength (λ) is given by:
\[ v = f \cdot \lambda \]
Rearranging this gives:
\[ \lambda = \frac{v}{f} \]

3. **Calculate the wavelength (λ)**:
Substitute the values of v and f into the equation:
\[ \lambda = \frac{220 \, \text{m/s}}{660 \, \text{Hz}} = \frac{220}{660} = \frac{1}{3} \, \text{m} \]

4. **Relate the length of the string to the number of loops**:
For a string clamped at both ends, the length (L) of the string is related to the wavelength (λ) by the number of loops (n):
\[ L = \frac{n \cdot \lambda}{2} \]
Here, n = 3.

5. **Calculate the length of the string (L)**:
Substitute the value of λ and n into the equation:
\[ L = \frac{3 \cdot \left(\frac{1}{3} \, \text{m}\right)}{2} = \frac{3}{6} \, \text{m} = 0.5 \, \text{m} = 50 \, \text{cm} \]

### Part (b): Write a suitable equation describing the motion.

1. **Identify the maximum amplitude**:
- Amplitude (A) = 0.5 cm = 0.005 m (convert to meters for standard units)

2. **Write the general equation for wave motion**:
The general equation for a wave can be expressed as:
\[ y(x, t) = A \sin\left(2\pi \left(\frac{x}{\lambda} - \frac{t}{T}\right)\right) \]
where T is the period of the wave.

3. **Calculate the period (T)**:
The period (T) is the reciprocal of the frequency (f):
\[ T = \frac{1}{f} = \frac{1}{660} \, \text{s} \]

4. **Substitute the values into the wave equation**:
- Wavelength (λ) = 1/3 m
- Substitute A, λ, and T into the wave equation:
\[ y(x, t) = 0.005 \sin\left(2\pi \left(\frac{x}{\frac{1}{3}} - 660t\right)\right) \]
Simplifying gives:
\[ y(x, t) = 0.005 \sin\left(6\pi x - 1320\pi t\right) \]

### Final Answers:
- (a) Length of the string = 50 cm
- (b) Suitable equation describing the motion:
\[ y(x, t) = 0.005 \sin\left(6\pi x - 1320\pi t\right) \]