Question:
Two vibrating tuning forks produce progressive waves given by
\[
y_1=2\sin(500\pi t-ax)
\]
and
\[
y_2=2\sin(506\pi t-bx).
\]
The number of beats produced per minute is
Two vibrating tuning forks produce progressive waves given by
\[
y_1=2\sin(500\pi t-ax)
\]
and
\[
y_2=2\sin(506\pi t-bx).
\]
The number of beats produced per minute is
Show Hint
If a wave is written as
\[
y=A\sin(\omega t-kx),
\]
then
\[
f=\frac{\omega}{2\pi}.
\]
Beat frequency is
\[
|f_1-f_2|,
\]
and beats per minute are
\[
60|f_1-f_2|.
\]
Updated On: Jul 9, 2026
- \(360\)
- \(180\)
- \(60\)
- \(3\) \bigskip
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
For two waves of frequencies \(f_1\) and \(f_2\),
\[
\text{Beat frequency}
=
|f_1-f_2|.
\]
The number of beats per minute is
\[
60\,|f_1-f_2|.
\]
Step 1: Determine the frequencies of the two waves. Comparing \[ y_1=2\sin(500\pi t-ax) \] with \[ y=A\sin(\omega t-kx), \] we get \[ \omega_1=500\pi. \] Hence, \[ f_1=\frac{\omega_1}{2\pi} =\frac{500\pi}{2\pi} =250\,\text{Hz}. \] Similarly, \[ \omega_2=506\pi. \] Therefore, \[ f_2=\frac{\omega_2}{2\pi} =\frac{506\pi}{2\pi} =253\,\text{Hz}. \]
Step 2: Calculate the beat frequency. \[ f_b = |f_2-f_1|. \] \[ f_b = |253-250|. \] \[ f_b=3\,\text{Hz}. \]
Step 3: Find the number of beats per minute. \[ N = 60\times3. \] \[ N=180. \]
Step 4: Write the final answer. \[ \boxed{180} \] \[ \boxed{\text{Answer = (B)}} \]
Step 1: Determine the frequencies of the two waves. Comparing \[ y_1=2\sin(500\pi t-ax) \] with \[ y=A\sin(\omega t-kx), \] we get \[ \omega_1=500\pi. \] Hence, \[ f_1=\frac{\omega_1}{2\pi} =\frac{500\pi}{2\pi} =250\,\text{Hz}. \] Similarly, \[ \omega_2=506\pi. \] Therefore, \[ f_2=\frac{\omega_2}{2\pi} =\frac{506\pi}{2\pi} =253\,\text{Hz}. \]
Step 2: Calculate the beat frequency. \[ f_b = |f_2-f_1|. \] \[ f_b = |253-250|. \] \[ f_b=3\,\text{Hz}. \]
Step 3: Find the number of beats per minute. \[ N = 60\times3. \] \[ N=180. \]
Step 4: Write the final answer. \[ \boxed{180} \] \[ \boxed{\text{Answer = (B)}} \]
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