You have two tuning forks. One is a known frequency of 200 Hz. The second for his plate, there is a big frequency of five beats per second. What a piece of gum is placed on the second fork, the frequency increases three beats per second. What is the frequency of the second tuning fork?
sin w t + sin x t = 2 sin.5(w+x)t cos.5(w-x)t
beat frequency is
omega = 2 pi f = .5(w-x)
assume gum decreases x
thereby increasing w-x
so
w is our 200 Hz fork
w = 200*2 pi = 400 pi radians/s
w - x = 5*2 pi = 10 pi
so
x = 390 pi = 195 Hz
I suppose that is more than you wanted to know but that is where it comes from :)
To determine the frequency of the second tuning fork, we can use the concept of beats. Beats occur when two sound waves with slightly different frequencies interfere with each other, resulting in a periodic increase and decrease in the volume or intensity of the sound.
Given that the first tuning fork has a known frequency of 200 Hz and the second tuning fork produces five beats per second, we can conclude that the frequency of the second tuning fork is either slightly lower or higher than 200 Hz.
When a piece of gum is placed on the second tuning fork, the frequency increases by three beats per second. This means that the frequency of the modified second tuning fork is either 3 Hz higher than the original frequency (if the initial frequency was lower than 200 Hz) or 3 Hz lower than the original frequency (if the initial frequency was higher than 200 Hz).
To find the frequency, we can subtract or add 3 Hz to the known frequency of 200 Hz, depending on whether the original frequency was higher or lower. However, this information is not provided in the question, so we cannot determine the exact frequency of the second tuning fork without additional information.