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-BARSIC- [3]
3 years ago
14

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

f 9.50 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Assume that the speed of sound in air is 343 m/s, find (a) that frequency and (b) the tension in the wire.
Physics
1 answer:
Ksju [112]3 years ago
7 0

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m

The fundamental frequency of the wire is given by

f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N

The tension in the wire is 67.9064 N

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Answer:

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Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

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Answer:

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