Question

Two identical piano wires, kept under the same tension $T$ have a fundamental frequency of $600 \,Hz$. The fractional increase in the tension of one of the wires which will lead to occurrence of $6 $ beat's when both the wires oscillate together would be

• A. 0.01
• B. 0.02
• C. 0.03
• D. 0.04

Answer 1

The Correct Option is B

 To solve this problem, we need to understand the concept of beats in wave physics and how the frequency of a vibrating string is related to the tension in the string.

When two waves of slightly different frequencies interfere, they produce a phenomenon known as beats. The beat frequency is given by the absolute difference between the two frequencies:

\(f_{\text{beat}} = |f_1 - f_2|\)

According to the problem, two identical piano wires have the same initial tension \(T\), resulting in a fundamental frequency of \(f_1 = 600 \, \text{Hz}\).

If the tension of one wire is increased slightly, the frequency of that wire will also increase. Let's say the tension is increased to \(T + \Delta T\), resulting in a new frequency \(f_2\).

The formula for the frequency of a vibrating string is:

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

where \(L\) is the length of the string and \(\mu\) is the linear density. Since both wires are identical and initially under the same tension, we can use the relation for frequency dependent on tension:

\(\frac{f_2}{f_1} = \sqrt{\frac{T + \Delta T}{T}}\)

Given that the beats produced are 6 per second, the beat frequency is:

\(f_{\text{beat}} = 6 \, \text{Hz}\)

Therefore,

f_2:

• \(f_2 = 600 + 6 = 606 \, \text{Hz}\)
• \(f_2 = 600 - 6 = 594 \, \text{Hz}\) (but this will not happen as tension increase leads to frequency increase)

So, \(f_2 = 606 \, \text{Hz}\).

Substitute back:

\(\frac{606}{600} = \sqrt{\frac{T + \Delta T}{T}}\)

Squaring both sides:

\(\left(\frac{606}{600}\right)^2 = \frac{T + \Delta T}{T} \Rightarrow \frac{606^2}{600^2} = \frac{T + \Delta T}{T}\)

Simplifying:

\(\left(\frac{606}{600}\right)^2 = \frac{368436}{360000} \approx 1.02\)

So,

\(\frac{T + \Delta T}{T} \approx 1.02\)

Implying \(T + \Delta T = 1.02 T\)

Thus, \(\Delta T = 0.02 T\)

Therefore, the fractional increase in tension required is 0.02.