Acoustics and Frequency Question (easy)

AI Thread Summary
Two tuning forks, one at 384Hz, produce 7 beats per second when sounded together, indicating a frequency difference of 7Hz. When a clamp is added to the 384Hz fork, the beat frequency reduces to 5 beats per second. This suggests the second fork's frequency is lower than 384Hz. The calculations indicate the frequency of the other fork is 377Hz, derived from the difference in beats. Understanding the impact of the clamp on the fork's frequency is crucial to solving the problem.
sobo22
Messages
2
Reaction score
0
1. Two nearly identical forks one of which has a frequency of 384Hz produce 7 beats per second when they vibrate at the same time. A small clamp is placed on the 384Hz fork and then only 5 beats per second are heard. What is the frequency of the other fork?



Homework Equations



I honestly have no idea

The Attempt at a Solution



Well I know the answer is 377 and I've been trying to work backwards but the only thing I can think of is 5 +/- uncertainty then 2 for the change in beats so 7? But even then I have no idea why it would be 377 vs 381.. Can someone please help me out I'm so lost
 
Physics news on Phys.org
Look up the equation for beating (Wikipedia has it, for example).
 
hey thanks man. So would it be

f = f1 - f2

f2 = f1 - f

f2 = 384 - 7 (amount of beats) = 377?

what about the 7 to 5 beats thing, is that just irrelevant filler?
does someone mind explaining the concept of it? I am still lost

is it because 7 beats are heard so that means there is a difference of 7 hz?
 
At first there are 7 beats between the forks, so you know their frequency difference -- but do you actually know whether the second fork is higher or lower than 384 Hz?

What happens to the frequency when you attach a "small" clamp to the 384 Hz fork?
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top