Question

Two tuning forks having frequencies $450\mathrm{Hz}$and $448\mathrm{Hz}$ are tuned together. The time interval between successive instants of maximum intensity is

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{second}$
• B. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\text{second}$
• C. $2\text{second}$
• D. $1\text{second}$

Answer 1

The correct option is A

$\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\text{second}$

Step 1: Given data

First frequency of tuning fork$\left(f_1\right)=450\text{Hz}$

Second frequency of tuning fork$\left(f_2\right)=448\text{Hz}$

Step 2: To find

How much time is required to hear the minimum sound?

Step 3: Formula used

The time interval among two consecutive peaks or minima is called the beat period. It is additionally equivalent to the beat frequency reciprocal.

1. $\text{Beat period}=\left|\frac{1}{f_2-f_1}\right|$
2. $\mathrm{Time}\mathrm{period}=\mathrm{Beat}\mathrm{period}$

Step 4: Calculate the time required to hear the minimum sound

$$\begin{gathered} \mathrm{Beat}\mathrm{period}=\left|\frac{1}{448\mathrm{Hz}-450\mathrm{Hz}}\right| \\ =\left|-\frac{1}{2\mathrm{Hz}}\right| \\ =\frac{1}{2}\mathrm{seconds} \end{gathered}$$

$\text{Time period}=\frac{1}{2}\text{seconds}$

Hence, option (A) is correct.