How Does Tightening a Guitar String Affect Its Frequency?

AI Thread Summary
Tightening a guitar string increases its tension, which raises its frequency. Initially, the beat frequency between the tuning fork and the guitar string is 3 Hz. After tightening the string, the beat frequency decreases to 2 Hz, indicating that the guitar string's frequency has increased and is now closer to the tuning fork's frequency. The new frequency of the guitar string can be calculated by considering the initial frequency and the change in beat frequency. Understanding the relationship between tension, frequency, and beat frequency is crucial for solving this problem.
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[SOLVED] Assignment Question Help! - Tuning Fork/ frequency

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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If you want help you need to put some effort into the problem. Those are the rules. So, tell me something about the problem. What concepts apply? What formulas apply?
 
I think i almost have it, ill msg back when i get an answer
 
Cool! Good Luck.
 
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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