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Step-by-Step Solution
Step 1: Identify the Given Data
• Length of string 1, $l_1 = 0.516 \text{ m}$
• Length of string 2, $l_2 = 0.491 \text{ m}$
• Tension in each string, $T = 20 \text{ N}$
• Mass per unit length, $\mu = 0.001 \text{ kg/m}$
• We need to find the number of beats when both strings are sounded together.
Step 2: Recall the Frequency Formula for a Stretched String
For a string of length $l$, under tension $T$ and having mass per unit length $\mu$, the fundamental frequency $ \nu $ is given by:
$$ \nu = \frac{1}{2l} \sqrt{\frac{T}{\mu}} $$
Step 3: Calculate the Frequencies of the Two Strings
• Frequency of the first string, $ \nu_1 $:
$$ \nu_1 = \frac{1}{2 \times l_1}\sqrt{\frac{T}{\mu}}
= \frac{1}{2 \times 0.516}\sqrt{\frac{20}{0.001}} $$
• Frequency of the second string, $ \nu_2 $:
$$ \nu_2 = \frac{1}{2 \times l_2}\sqrt{\frac{T}{\mu}}
= \frac{1}{2 \times 0.491}\sqrt{\frac{20}{0.001}} $$
Step 4: Determine the Number of Beats
Beats per second is the absolute difference between the two frequencies:
$$ \text{Number of beats} = |\nu_1 - \nu_2| $$
On computation, this difference comes out to be approximately 7.
Step 5: State the Final Answer
Therefore, the number of beats heard when both strings are vibrated simultaneously is 7.
