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Answer to Question #1836 in Mechanics | Relativity for nirmala

Question #1836
A wave is described by the following equation:
y = (1.0 nm) sin 2π[t/.01s-x/2cm]
<br>(i) Calculate the time period and the wavelength of the wave.
<br>(ii) Obtain the expression for the velocity of the particles.
<br>(iii) Calculate the speed of the particles at x = 3.0 cm, 5.0 cm and 7.0 cm at t = 0.01 s.
<br>(iv) Calculate speed of the particles at x = 1.0 cm, at t = 0.011 s, 0.012 s and 0.013s.

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Expert's answer
The time period& of the wave:& T = 2&pi; / (2&pi;/0.01) = 100 s,& the wavelength is& &lambda; = 2&pi; / (2&pi;/2) = 2 cm.
The velocity is dy/dt = 200&pi; cos 2&pi;[t/.01-x/2] [nm/s]
& V(x=3.0cm, t= 0.01 s) = 200&pi; cos 2&pi;(.01/.01-3/2) = -627.999 nm/s
& V(x=5.0cm, t= 0.01 s) =& -627.992 nm/s
& V(x=7.0cm, t= 0.01 s) =& -627.980 nm/s

& V(x=1.0cm, t= 0.011 s) = 200&pi; cos 2&pi;(.011/.01-1/2) = -508.767 nm/s
& V(x=1.0cm, t= 0.012 s) = 200&pi; cos 2&pi;(.012/.01-1/2) = -195.394 nm/s
& V(x=1.0cm, t= 0.013 s) = 200&pi; cos 2&pi;(.013/.01-1/2) = 192.540 nm/s

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Comments

Assignment Expert
09.03.11, 16:05

This sign is not significant for the solution.

nirmala
09.03.11, 13:39

A wave is described by the following equation: y = (1.0 nm) sin 2π[t/.01s-x/2cm]
(i) Calculate the time period and the wavelength of the wave.

standard wave form is y= a sin (kx-wt) and my doubt is we should consider (-) sing also while comparing time period and wavelength

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