What is the expected period of beats formed when two tuning forks sound simultaneously and have the frequencies f1= 250Hz and 242Hz?
you get the sum, and the difference. 8hz, 592hz It will be difficult to hear 8 hz.
then after that 1/592Hz?
you will hear definitely 592hz.
so the period is 1/592 sec
period is 1/f, right?
so if I am looking for period I need to 1/592Hz....which is1.689*10^-3 s, which is very fast
you will at least hear it....
Sorry If the comment above was repeated...Something happened...
so the period 1.689*10^3s
thank you
To find the expected period of beats formed when two tuning forks sound simultaneously, with frequencies f₁ = 250 Hz and f₂ = 242 Hz, we need to calculate the difference between the two frequencies.
The formula to find the period of beats is:
Period of beats (Tᵦ) = 1 / (Δf)
Where Δf is the difference in frequencies between the two tuning forks.
In this case, the difference in frequencies is Δf = |f₁ - f₂| = |250 Hz - 242 Hz| = 8 Hz.
Now we can substitute the value of Δf into the formula to calculate the period of beats:
Tᵦ = 1 / 8 Hz ≈ 0.125 seconds
Therefore, the expected period of beats formed when these two tuning forks sound simultaneously is approximately 0.125 seconds.