How Does Changing String Tension Affect Frequency in a Physics Problem?

AI Thread Summary
Changing the tension of a string affects its frequency, causing it to vibrate at a higher pitch. In the discussion, two strings initially vibrate at 260 Hz, but when one string is tightened, it produces three beats per second with the other string. The formula Fbeat = F1 - F2 suggests that the new frequency of the tightened string could be calculated as 263 Hz, assuming it is greater than the original frequency. However, isolating the second frequency results in 257 Hz, leading to confusion about the correct new frequency. The key takeaway is that increased tension raises frequency, but the relationship between the two frequencies must be carefully considered to determine the accurate value.
wentwork882
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Two strings are adjusted to vibrate at 260 Hz. One string is tightened then 3 beats per second is head when the string vibrate at the same time. What is the new frequency of the tightened string.
I am using Fbeat=F1-F2; so wouldn't the answer just be 263 Hz
 
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Well, can you explain why 263 Hz, and not 257 Hz?
 
The tension increased so the frequency must be greater than the F1. But when F2 is isolated the result would be 260-3 = 257hz
 
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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