Question

Two identical wires have a fundamental frequency of $100 \text{ Hz}$ when kept under the same tension. If the tension of one of the wires is increased by $21\%$, the number of beats produced is

• A. $11$
• B. $10$
• C. $9$
• D. $8$

Hint:

Square root values are common in physics: $\sqrt{1.21} = 1.1$, $\sqrt{1.44} = 1.2$, and $\sqrt{1.69} = 1.3$. Recognizing these allows for much faster mental math.

Answer 1

The Correct Option is B

Beat frequency is the absolute difference between two frequencies. We need to find the new frequency of the wire after its tension is increased. Step 1: Identify the Frequency-Tension Relationship The frequency ($n$) of a vibrating string is proportional to the square root of its tension ($T$): $$n \propto \sqrt{T}$$

Step 2: Calculate the New Frequency ($n'$) The initial tension is $T$ and the initial frequency is $n = 100 \text{ Hz}$. The new tension $T'$ is $T + 0.21T = 1.21T$. $$\frac{n'}{n} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{1.21T}{T}} = \sqrt{1.21} = 1.1$$ $$n' = 1.1 \times n = 1.1 \times 100 = 110 \text{ Hz}\lt strong\gt Step 3: Calculate Beat Frequency\lt /strong\gt \text{Beats} = n' - n = 110 - 100 = 10 \text{ beats/s}$$