Asked by Anonymous
a wire under tension vibrates with a fundamental frequency of 256Hz. what would be the fundamental frequency if the wire were half as long, twice as thick, and under 1/4 tension?
Answers
Answered by
drwls
The speed of transverse waves is the squareroot of (T/p), where p is the density per unit length, and T is the tension foorce. The density per unit length is 4x higher for the shorter, fatter wire, and T is 1/4 as high, to the sound speed is 1/4 as large.
The wavelength of the fundamental is 1/2 as long due to the shorter length.
The fundamental frequency scales with (sound speed)/(wavelength), and gets multiplied by (1/4)/(1/2) = 1/2
The wavelength of the fundamental is 1/2 as long due to the shorter length.
The fundamental frequency scales with (sound speed)/(wavelength), and gets multiplied by (1/4)/(1/2) = 1/2
Answered by
Katerina
f1=v1/L1=(1/L1)(T1/μ1)^1/2
f2=v2/L2=(1/L2)(T2/μ2)^1/2
then
f2/f1=(L1/L2)[T2μ1/(T1μ2)]^1/2
where
L2=1/2L1
T2=1/4T1
μ2=ρS2=4ρS1=4μ1
then
f2=1/2f1
f2=128 hz
f2=v2/L2=(1/L2)(T2/μ2)^1/2
then
f2/f1=(L1/L2)[T2μ1/(T1μ2)]^1/2
where
L2=1/2L1
T2=1/4T1
μ2=ρS2=4ρS1=4μ1
then
f2=1/2f1
f2=128 hz
Answered by
june
A WIRE UNDER TENSION VIBRATES WITH A FUNDAMENTAL FREQUENCY OF 256 HZ. WHAT WOULD BE
THE FUNDAMENTAL FREQUENCY IF THE WIRE WERE HALF AS LONG, TWICE AS THICK AND UNDER
ONE-FOURTH THE TENSION
Given: f1 = 256 Hz
L1 = ½ L1
T2 = ¼ T1
Find: f2
THE FUNDAMENTAL FREQUENCY IF THE WIRE WERE HALF AS LONG, TWICE AS THICK AND UNDER
ONE-FOURTH THE TENSION
Given: f1 = 256 Hz
L1 = ½ L1
T2 = ¼ T1
Find: f2
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