What is the period and amplitude for #I(t) =120 sin (10pix - pi/4)#?

1 Answer
Manisit Das
Jun 5, 2015

A general time-dependent wave function can be represented in the following form:

#y = A*sin(kx-omegat)#

where,
#A# is amplitude
#omega = (2pi)/T# where #T# is time period
#k = (2pi)/lamda# where #lamda# is the wavelength

So, comparing with the given equation #I(t) =120 sin (10pix - pi/4)#, we can find:

Amplitude (#A#) = 120

Now, your supplied equation has no t- dependent parameter in the sine function, whereas the L.H.S. clearly indicates it is a time-dependent function [#I(t)#]. So, this is impossible!

Probably, your equation was supposed to be #I(t) =120 sin (10pix - pi/4t)#

Under that condition,
#omega = pi/4#
#=> pi/4 = (2pi)/T#
#=> T = 8# units

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