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A stretched string of length L is observed to vibrate in five equal segments when driven by a 709-Hz oscillat?
A stretched string of length L is observed to vibrate in five equal segments when driven by a 709-Hz oscillator. What oscillator frequency will set up a standing wave so that the string vibrates in two segments?
____Hz
(b) Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 4,000 Hz, find the distance between adjacent antinodes for this mode of vibration.
____cm
1 Answer
- JullyWumLv 78 years agoFavorite Answer
a) The frequency of the first wave (at f1 = 709 Hz) is greater than at the second frequency (f2) ..
f1 produces 2½ waves (5, ½-wave segments) in length L and f2 produces 1 wave in length L .. .. so, f1 = 2.50 x f2
f2 = f1/2.50 = 709 / 2.50 .. .. ►f2 = 283.60 Hz
(b) Wavelength at 4000 Hz and sound velocity 340 m/s ..
λ = v/f = 340 / 4000 .. λ = 0.085 m
Adjacent anti-nodes are separated by ½λ ..
.. d = 0.085m / 2 .. ►d = 0.0425 m (4.25 cm)
