Question 243: Waves

Question

A tuning fork gives 5 beats per second with 40 cm length of sonometer wire. If the length of the wire is shortened by 1 cm , the number of beats is still the same. The frequency of the fork is

Accepted Answer

Step 1: Formula for Wire Frequency
The frequency ($\mathrm{f}$) of a vibrating wire is given by $$\mathrm{f} = \frac{1}{2l} \sqrt{\frac{T}{m}}$$ where $l$ is the length of the wire, $T$ is the tension, and $m$ is the mass per unit length.
Step 2: Frequency and Length Relationship
This formula shows that frequency is inversely proportional to length, so when the length decreases, the frequency increases. We can write: $$\mathrm{f} \propto \frac{1}{l}$$ or $$\mathrm{f} \cdot l = 	ext{constant}$$
Step 3: Naming the Frequencies
Let $f$ be the frequency of the tuning fork. Let $f_1$ be the wire's frequency with length $l_1 = 40$ cm, and $f_2$ be the wire's frequency with length $l_2 = 39$ cm (after shortening by 1 cm).
Step 4: Beats Before and After Change
Before reducing the length, the number of beats per second is: $f - f_1 = 5$
After reducing the length, the number of beats per second is: $f_2 - f = 5$
Step 5: Using the Proportionality
Because $\mathrm{f} \cdot l = 	ext{constant}$, we have: $$f_1 l_1 = f_2 l_2$$
Step 6: Substitute Values
$f_1 = f - 5$, $f_2 = f + 5$, $l_1 = 40$ cm, $l_2 = 39$ cm.
Plug in these values: $$(f - 5) \cdot 40 = (f + 5) \cdot 39$$
Step 7: Solve for $f$
Expand both sides: $$40f - 200 = 39f + 195$$
Rearrange the terms so $f$ is on one side: $$40f - 39f = 195 + 200$$
Simplify: $$f = 395\,\mathrm{Hz}$$

Answer

390 Hz

Answer

395 Hz

Answer

400 Hz

Answer

405 Hz

Waves

A tuning fork gives 5 beats per second with 40 cm length of sonometer wire. If the length of the wire is shortened by 1 cm , the number of beats is still the same. The frequency of the fork is

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