# 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.

<br>

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.

<br>

Expert's answer

The time period& of the wave:& T = 2π / (2π/0.01) = 100 s,& the wavelength is& λ = 2π / (2π/2) = 2 cm.

The velocity is dy/dt = 200π cos 2π[t/.01-x/2] [nm/s]

& V(x=3.0cm, t= 0.01 s) = 200π cos 2π(.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) = 200π cos 2π(.011/.01-1/2) = -508.767 nm/s

& V(x=1.0cm, t= 0.012 s) = 200π cos 2π(.012/.01-1/2) = -195.394 nm/s

& V(x=1.0cm, t= 0.013 s) = 200π cos 2π(.013/.01-1/2) = 192.540 nm/s

The velocity is dy/dt = 200π cos 2π[t/.01-x/2] [nm/s]

& V(x=3.0cm, t= 0.01 s) = 200π cos 2π(.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) = 200π cos 2π(.011/.01-1/2) = -508.767 nm/s

& V(x=1.0cm, t= 0.012 s) = 200π cos 2π(.012/.01-1/2) = -195.394 nm/s

& V(x=1.0cm, t= 0.013 s) = 200π cos 2π(.013/.01-1/2) = 192.540 nm/s

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## Comments

Assignment Expert09.03.11, 16:05This sign is not significant for the solution.

nirmala09.03.11, 13:39A 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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