Courses for Kids
Free study material
Offline Centres
Store Icon

Two tuning forks when surrounded together produce $6$ beats per second. The first fork has the frequency $3\% $ higher than a standard one and the second has the frequency $2\% $ less than the standard fork. The frequencies (in $Hz$) of the forks are:
A. $126.3,116.7$
B. $162.3,161.7$
C. $136.2,137.2$
D. ($123.6,117.6$

Last updated date: 17th Jun 2024
Total views: 402.9k
Views today: 4.02k
402.9k+ views
- We can solve this problem by using the concept of beats of tuning forks. We should know about wave- interference.
- We know that a tuning fork is a two pronged metal device that vibrates when struck.
- When the waves of sound are in the same phase they lead to constructive interference and when they are 180 degrees out of phase, they lead to destructive interference.
- A beat is an interference pattern of sounds of different frequencies.

Complete step by step solution:

In the same medium when two waves of sounds meet while travelling on intersecting or coincident paths, a net effect of combination of two or more waves will arise. It is termed as interference.

Let us consider the frequency of two tuning forks $\mathop n\nolimits_1 $ and$\mathop n\nolimits_2 $. Here the beat between the sound waves emitted from the tuning fork is $6$.
Therefore,$\mathop n\nolimits_1 - \mathop n\nolimits_2 = 6$ (i)
[As beat is equal to the difference of frequencies of two tuning forks here]

Now, the first fork has the frequency of $3\% $ higher than a standard one.
Let us assume, the frequency of standard one is equal to $n$.

Therefore, $\mathop n\nolimits_1 = (n + n \times \dfrac{3}{{100}}) = n \times (1 + \dfrac{3}{{100}}) = \dfrac{{103}}{{100}}n$
Similarly, for the second fork, it has the frequency $2\% $ less than the standard fork.
Therefore, $\mathop n\nolimits_2 = (n - n \times \dfrac{2}{{100}}) = n \times (1 - \dfrac{2}{{100}}) = \dfrac{{98}}{{100}}n$

Now, putting the values of $\mathop n\nolimits_1 $ and$\mathop n\nolimits_2 $ in the equation (i)
$\mathop n\nolimits_1 - \mathop n\nolimits_2 = 6$
$ \Rightarrow $ $\dfrac{{103n}}{{100}} - \dfrac{{98n}}{{100}} = 6$

$ \Rightarrow $ $\dfrac{{5n}}{{100}} = 6$

$ \Rightarrow $ $n = \dfrac{{100 \times 6}}{5} = 120$

Therefore, $n = 120Hz$.

Now, the frequency of first tuning fork $ \Rightarrow \mathop n\nolimits_1 = \dfrac{{103n}}{{100}} = \dfrac{{103 \times 120}}{{100}} = 123.6$

And the frequency of the second tuning fork $ \Rightarrow \mathop n\nolimits_2 = \dfrac{{98n}}{{100}} = \dfrac{{98 \times 120}}{{100}} = 117.6$

Therefore $\mathop n\nolimits_1 = 123.6Hz$ and $\mathop n\nolimits_2 = 117.6Hz$.

Hence the correct option is (D)

- All the waves possess the property of interference.
- There are two types of interference. One is constructive interference and the other is destructive interference.
- Beats cannot be heard if the difference of the frequencies is more than $10Hz$.
- Beats are dependent on the amplitude of the sound waves. Beats cannot be heard properly if the difference of the amplitude is high.