Each of the two strings of length $51.6 ext{ cm}$ and $49.1 ext{ cm}$ are tensioned separately by $20 ex

Question

Each of the two strings of length $51.6 	ext{ cm}$ and $49.1 	ext{ cm}$ are tensioned separately by $20 	ext{ N}$ force. Mass per unit length of both the strings is same and equal to $1 	ext{ g m}^{-1}$. When both the strings vibrate simultaneously the number of beats is:

Accepted Answer

1. **Identify the formula for fundamental frequency:** The frequency of a vibrating string is given by $f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$, where $L$ is the length, $T$ is the tension, and $\mu$ is the mass per unit length .
2. **Calculate the wave speed ($v$):** Given tension $T = 20 	ext{ N}$ and linear mass density $\mu = 1 	ext{ g m}^{-1} = 10^{-3} 	ext{ kg m}^{-1}$.
   $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 	ext{ m/s}$$
3. **Calculate the frequencies of both strings:**
   - For the first string ($L_1 = 51.6 	ext{ cm} = 0.516 	ext{ m}$):
     $$f_1 = \frac{v}{2L_1} = \frac{141.42}{2 	imes 0.516} = \frac{141.42}{1.032} \approx 137.03 	ext{ Hz}$$
   - For the second string ($L_2 = 49.1 	ext{ cm} = 0.491 	ext{ m}$):
     $$f_2 = \frac{v}{2L_2} = \frac{141.42}{2 	imes 0.491} = \frac{141.42}{0.982} \approx 144.01 	ext{ Hz}$$
4. **Calculate the beat frequency:** The beat frequency is the difference between the two frequencies .
   $$f_{	ext{beat}} = |f_2 - f_1| = |144.01 - 137.03| = 6.98 \approx 7$$

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