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
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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.
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 $T + 0.21T = 1.21T$.
$$\frac{n'}{n} = \sqrt{\frac{T'}{T}} = \sqrt{\frac{1.21T}{T}} = \sqrt{1.21} = 1.1$$
$$n' = 1.1 \times n = 1.1 \times 100 = 110 \text{ Hz}\lt strong\gt Step 3: Calculate Beat Frequency\lt /strong\gt \text{Beats} = n' - n = 110 - 100 = 10 \text{ beats/s}$$