Solving the Sound Wave Problem of Two Identical Violin Strings

AI Thread Summary
Two identical violin strings, both initially tuned to 440.0 Hz, produce a beat frequency of 1.5 beats per second when one string is retuned. The highest possible fundamental frequency of the retuned string is 441.5 Hz, while the lowest is 438.5 Hz, reflecting the beat frequency's influence. The fractional change in tension can be calculated using the relationship between frequency and tension, specifically that frequency is proportional to the square root of tension. If the frequency increases, the tension change ratio is derived from the formula f_2 = sqrt(T2/T1) * f_1. The hint provided indicates that the fractional change in tension can be expressed as δT/T = 2 * δf/f, guiding the calculation of tension adjustments.
Lakers08
Messages
7
Reaction score
0
Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 Hz. One of the strings is retuned by adjusting its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously.

-What is the highest possible fundamental frequency of the retuned string?
-What is the lowest possible fundamental frequency of the retuned string?
-By what fractional amount was the string tension changed if it was increased?
-By what fractional amount was the string tension changed if it was decreased?

please help me get started I am totally stomped, thanks
 
Physics news on Phys.org
HINT: The beat frequency is the difference between the two frequencies.
 
thanks a lot, lol for some reason i had the equation with a plus sign, thanks for the hint if it wasent for you i would be stuck here all night.
 
ummm just ran into another problem in this question, for the third part "by what fractional amount was the string changed?"

Iam using that the fundamental frequency of a string is proporional to the velocity of the waves which is proportional to the square root of the Tension in the string:

for example : f_2 = sqrt(T2/T1)*f_1

I am plugging in the frequency that i got for part a? for example 440 will go on f_1 and 442 will go on f_2 and then i solve for the ratio of T2/T1 but somehow I am missing something or I am doing this wrong, someone please help
 
HINT:

\frac {\delta T}{T} = 2 \frac {\delta f}{f}
 
thanks a lot
 
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}...
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...
Back
Top