What is the amplitude and period of #x=asin(nt)+bcos(nt)#?

1 Answer
Shwetank Mauria
Apr 6, 2017

Amplitude is #sqrt(a^2+b^2)# and period is #(2pi)/n#.

Explanation:

As #x=asin(nt)+bcos(nt)#, we can write ir as #x=bcos(nt)+asin(nt)#

Now let #cosalpha=a/sqrt(a^2+b^2)# and #sinalpha=b/sqrt(a^2+b^2)#

Observe that as #|a/sqrt(a^2+b^2)|<1# and #|b/sqrt(a^2+b^2)|<1#, it is possible as #cos^2alpha+sin^2alpha=1#

Then #x=sqrt(a^2+b^2)[cos(nt)cosalpha+sin(nt)sinalpha]#

= #sqrt(a^2+b^2)cos(nt-alpha)#

Hence amplitude is #sqrt(a^2+b^2)# and period is #(2pi)/n#.

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