How Does Tightening a Guitar String Affect Beat Frequency?

AI Thread Summary
Tightening a guitar string increases its frequency, which affects the beat frequency heard when played alongside a tuning fork. Initially, a beat frequency of 3 Hz indicates a difference between the string's frequency and the tuning fork's frequency. After tightening the string, the beat frequency decreases to 2 Hz, suggesting the string's frequency has moved closer to that of the tuning fork. To find the new frequency of the guitar string, one must consider that the original frequency was 3 Hz higher than the tuning fork's frequency, and the new frequency is now only 2 Hz higher. Therefore, the new frequency of the guitar string can be calculated accordingly.
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[SOLVED] Physics Assignment Question Help!

I've talked to numerous people who can not figure out this question, so if anyone has any ideas on how to do this question please help me!:confused:

A tuning fork is struck and held next to a vibrating guitar string, and beats of frequency 3 Hz are heard. The guitar string is tightened slightly, and the beat frequency decreases to 2 Hz. What is the new frequency of the guitar string? Explain your reasoning.
 
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Check your other post. Please do not double post problems.
 
Yeah i know I accidently posted it into the wrong section and then a moderator i think moved it here? not really sure I am knew, so i thought it was deleted, so i posted a new one in the homework section
 
No problem. It's just easier to help with homework problems we don't have a problem being solved more than once. Don't worry about it.
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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