Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

Step-by-Step Solution

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 .
Calculate the wave speed ( $v$ ): Given tension $T = 20 \text{ N}$ and linear mass density $\mu = 1 \text{ g m}^{-1} = 10^{-3} \text{ kg m}^{-1}$ . $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{10^{-3}}} = \sqrt{20000} = 100\sqrt{2} \approx 141.42 \text{ m/s}$
Calculate the frequencies of both strings:

For the first string ( $L_1 = 51.6 \text{ cm} = 0.516 \text{ m}$ ): $f_1 = \frac{v}{2L_1} = \frac{141.42}{2 \times 0.516} = \frac{141.42}{1.032} \approx 137.03 \text{ Hz}$
For the second string ( $L_2 = 49.1 \text{ cm} = 0.491 \text{ m}$ ): $f_2 = \frac{v}{2L_2} = \frac{141.42}{2 \times 0.491} = \frac{141.42}{0.982} \approx 144.01 \text{ Hz}$

Calculate the beat frequency: The beat frequency is the difference between the two frequencies . $f_{\text{beat}} = |f_2 - f_1| = |144.01 - 137.03| = 6.98 \approx 7$

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