For the harmonic travelling wave y = 2 cos 2π (10t–0.0080x + 3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of,
(a) 4 m
(b) 0.5 m
(c) ![]()
(d)
(at a given instant of time)
(e) What is the phase difference between the oscillation of a particle located at x = 100cm, at t = T s and t = 5 s?
Given:
The equation of the wave y = 2cos[2π(10t-0.0080x+3.5)]
(a) For the wave separated by a distance of 4 m the equation of vibration will be,
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or ![]()
and the phase
![]()
Comparing this equation to the original equation
![]()
and the phase
![]()
the phase difference therefore
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(b) For a separation of 0.5 m, proceeding as above,
![]()
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and the phase
![]()
Comparing this equation to the original equation its phase
![]()
the phase difference therefore
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(c) The wavelength of the equation can be found by comparing the equation to the general equation for the wave,
![]()
we have, k = 0.016π and ω =20π. We know that k is related to the wavelength as
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where λ is the wavelength. Therefore for k = 0.016π we have
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For a separation of
or 0.625 m,
![]()
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and the phase
![]()
Comparing this equation to the original equation its phase
![]()
the phase difference therefore
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(d) For a separation of
or 0.938 m, proceeding as above
![]()
![]()
and the phase
![]()
Comparing this equation to the original equation its phase
![]()
the phase difference therefore
![]()
(e) At x = 100 cm the equation of the wave becomes,
![]()
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We know that ω is and time period T are related as
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but ω = 20π,
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At t = T= 0.1 s the equation of the wave is given by,
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and the phase
![]()
At t = 5 s the equation of the wave is given by,
![]()
and the phase
![]()
the phase difference therefore
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