Question

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

• A. 7
• B. 8
• C. 3
• D. 5

Answer 1

The Correct Option is A

To find the number of beats produced when the two strings vibrate simultaneously, we need to calculate the frequencies of both strings and then find the difference between them.

The frequency \( f \) of a vibrating string is given by the formula:

f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}

• L is the length of the string.
• T is the tension in the string.
• \mu is the mass per unit length.

Given in the problem:

• String lengths: L_1 = 51.6\, \text{cm} = 0.516\, \text{m} and L_2 = 49.1\, \text{cm} = 0.491\, \text{m}
• Tension: T = 20\, \text{N}
• Mass per unit length: \mu = 1\, \text{g/m} = 0.001\, \text{kg/m}

Calculate the frequencies of both strings:

1. Frequency of the first string:

f_1 = \frac{1}{2 \times 0.516} \sqrt{\frac{20}{0.001}} = \frac{1}{1.032} \times \sqrt{20000}

f_1 \approx \frac{1}{1.032} \times 141.42 = \frac{141.42}{1.032} \approx 137.07\, \text{Hz}

2. Frequency of the second string:

f_2 = \frac{1}{2 \times 0.491} \sqrt{\frac{20}{0.001}} = \frac{1}{0.982} \times \sqrt{20000}

f_2 \approx \frac{1}{0.982} \times 141.42 = \frac{141.42}{0.982} \approx 144.19\, \text{Hz}

Now, calculate the number of beats:

The number of beats per second is the absolute difference between the two frequencies:

\text{Beats} = |f_1 - f_2| = |137.07 - 144.19| = 7.12

Rounding to the nearest whole number, we find that the number of beats is approximately 7.

Conclusion: The correct answer is 7.