A 250 kg weight is hanging from the ceiling by a thin copper wire. In its fundam

Question

A 250 kg weight is hanging from the ceiling by a thin copper wire. In its fundamental mode, this wire vibrates at the frequency of concert A (440 Hz). You then increase the temperature of the wire by 40 C

A 250 kg weight is hanging from the ceiling by a thin copper wire. In its fundamental mode, this wire vibrates at the frequency of concert A (440 Hz). You then increase the temperature of the wire by 40 C

a) By how much will the fundamental frequency change? (answer in Hz)

b)Will it increase or decrease?

c)By what percentage will the speed of a wave on the wire change?

d)By what percentage will the wavelength of the fundamental standing wave change?

e)Will wavelenght increase or decrease?

Explanation / Answer

Assuming linear thermal expansion, heating increases the length of the wire:
L = L(1 + T)
or
(L /L) = 1 + T

for copper
= 17×10K¹
So the length increases due to heating by a factor of
(L /L) = 1 + 17×10K¹40K
= 1.00068

(a)
The fundamental frequency of vibrating string is given by:
f = (T/µ) = ( T / (4µL²) )
where L length of the wire
T tension of the string
µ linear mass of string
Ignoring the small mass of the wire, the tension is determined by the weight force of the mass hanging on it. It is force divided by area of action:
T = Mg /A
(A is the cross-sectional area of the wire).
Actually the tension does not change due to thermal expansion.
The linear mass is mass of wire divided by its length
µ = m/L
Since m remains unchanged, µ changes due to heating.
Hence:
f = ( T / (4mL) )
In other words the frequency of constant mass constant tension wire is inversely proportional to the square root of the length.
So
f/f = ( L/L )
=>
f = f( L/L ) = f / ( L/L )
= 440Hz / ( 1.00068 )
= 439.85Hz

(b) Ofcourse it will decrease.

(c)
speed of on wave on a vibrating string is:
v = (T/µ) = ( TL /m)
That means the speed of wave is proportional to the square root of the length.
Therefore
v/v = (L/L)
To the relative change of velocity is
(v - v)/v = v/v - 1 = (L/L) - 1
= (1.00068) - 1
= 3.4×10
= 0.034%
Speed of wave increases by 0.034%

(d)
In fundamental mode wavelength is one half of the length of the wire:
= L/2
So
/ = L/L
To the relative change of wave length is
( - )/ = / - 1 = (L/L) - 1
= 1.00068) - 1
= 6.8×10
= 0.068%
(e)
Wavelength increases by 0.068%

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