
The frequency of a tuning fork is five percent greater than that of standard frequency k. Another tuning fork is three percent less than that of k. When both the forks are tuned together, four beats are produced. Find the frequencies of both the forks.
$\begin{align}
& a)52.5,48.5Hz \\
& b)57.5,47.5Hz \\
& c)40.5,42.5Hz \\
& d)none \\
\end{align}$
Answer
570.9k+ views
Hint: Number of beats can be calculated as the difference between the frequencies of the two given forks. As the standard frequency is given, write the frequencies of two forks in terms of standard frequency and find the value of standard frequency. Next, we can find out individual frequencies of the forks.
Formula used: beats=${{f}_{a}}-{{f}_{b}}$
Complete step by step answer:
Let the frequencies of two tuning forks be ${{f}_{a}},{{f}_{b}}$ respectively. Now, it is given in the question that the frequency of the first tuning fork is five percent greater than standard frequency. The second frequency is three percent less than standard frequency. Therefore, the value of frequencies will be ${{f}_{a}}=\dfrac{105}{100}K,{{f}_{b}}=\dfrac{97}{100}K$.
Now, we can calculate the beats as the difference between the given two frequencies. That is,
$\begin{align}
& 4=\dfrac{105}{100}K-\dfrac{97}{100}K \\
& K=\dfrac{400}{8} \\
& K=50Hz \\
\end{align}$
As we now found the value of standard frequency,
The values of both the frequencies will be.
$\begin{align}
& {{f}_{a}}=\dfrac{105}{100}(50) \\
& {{f}_{a}}=52.5Hz \\
& {{f}_{b}}=\dfrac{97}{100}(50) \\
& {{f}_{b}}=48.5Hz \\
\end{align}$
So, the correct answer is “Option A”.
Additional Information: Beat frequency is always equal to the difference between the frequency of the two notes that interfere to produce the beats. Beat frequencies are hard to distinguish. The beat period or period of beats is the time interval between two successive maxima or minima. It can also be stated as the time difference between waxing and waning. Number of beats heard per second is called the frequency of beats.
Note: A beat frequency can never be negative. The beat frequency is the absolute difference between the given frequencies. So, a negative beat frequency will never make any sense. When two frequencies become closer, the beats will eventually decrease and slowly disappear when they turn identical.
Formula used: beats=${{f}_{a}}-{{f}_{b}}$
Complete step by step answer:
Let the frequencies of two tuning forks be ${{f}_{a}},{{f}_{b}}$ respectively. Now, it is given in the question that the frequency of the first tuning fork is five percent greater than standard frequency. The second frequency is three percent less than standard frequency. Therefore, the value of frequencies will be ${{f}_{a}}=\dfrac{105}{100}K,{{f}_{b}}=\dfrac{97}{100}K$.
Now, we can calculate the beats as the difference between the given two frequencies. That is,
$\begin{align}
& 4=\dfrac{105}{100}K-\dfrac{97}{100}K \\
& K=\dfrac{400}{8} \\
& K=50Hz \\
\end{align}$
As we now found the value of standard frequency,
The values of both the frequencies will be.
$\begin{align}
& {{f}_{a}}=\dfrac{105}{100}(50) \\
& {{f}_{a}}=52.5Hz \\
& {{f}_{b}}=\dfrac{97}{100}(50) \\
& {{f}_{b}}=48.5Hz \\
\end{align}$
So, the correct answer is “Option A”.
Additional Information: Beat frequency is always equal to the difference between the frequency of the two notes that interfere to produce the beats. Beat frequencies are hard to distinguish. The beat period or period of beats is the time interval between two successive maxima or minima. It can also be stated as the time difference between waxing and waning. Number of beats heard per second is called the frequency of beats.
Note: A beat frequency can never be negative. The beat frequency is the absolute difference between the given frequencies. So, a negative beat frequency will never make any sense. When two frequencies become closer, the beats will eventually decrease and slowly disappear when they turn identical.
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