Question:
The frequency of a tuning fork P is 1.5% more than the frequency of tuning fork Q and the frequency of another tuning fork R is 2.5% less than the frequency of tuning fork Q. If 8 beats are produced per second when P and R are vibrated together, then the frequency of tuning fork R is
The frequency of a tuning fork P is 1.5% more than the frequency of tuning fork Q and the frequency of another tuning fork R is 2.5% less than the frequency of tuning fork Q. If 8 beats are produced per second when P and R are vibrated together, then the frequency of tuning fork R is
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Beat frequency:
\[
f_b=|f_1-f_2|
\]
Always convert percentage increase/decrease into decimal form before calculation.
Updated On: Jun 17, 2026
- $203\text{ Hz}$
- $195\text{ Hz}$
- $200\text{ Hz}$
- $187\text{ Hz}$
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The Correct Option is B
Solution and Explanation
Concept:
Beat frequency is equal to the absolute difference between the frequencies of two sound sources.
\[
f_b=|f_P-f_R|
\]
Step 1: Express the frequencies in terms of \(f_Q\).
Let the frequency of tuning fork \(Q\) be \(f\). Then, \[ f_P=f+1.5%f \] \[ f_P=1.015f \] Similarly, \[ f_R=f-2.5%f \] \[ f_R=0.975f \]
Step 2: Use the beat frequency relation.
Given, \[ f_b=8Hz \] Therefore, \[ 1.015f-0.975f=8 \] \[ 0.04f=8 \] \[ f=200Hz \]
Step 3: Calculate the frequency of tuning fork R.
\[ f_R=0.975(200) \] \[ f_R=195Hz \] Thus, the frequency of tuning fork \(R\) is \[ \boxed{195Hz} \]
Step 1: Express the frequencies in terms of \(f_Q\).
Let the frequency of tuning fork \(Q\) be \(f\). Then, \[ f_P=f+1.5%f \] \[ f_P=1.015f \] Similarly, \[ f_R=f-2.5%f \] \[ f_R=0.975f \]
Step 2: Use the beat frequency relation.
Given, \[ f_b=8Hz \] Therefore, \[ 1.015f-0.975f=8 \] \[ 0.04f=8 \] \[ f=200Hz \]
Step 3: Calculate the frequency of tuning fork R.
\[ f_R=0.975(200) \] \[ f_R=195Hz \] Thus, the frequency of tuning fork \(R\) is \[ \boxed{195Hz} \]
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