Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
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Chapter 5, Problem 5.48P
To determine

Prove the formulas given.

Expert Solution & Answer
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Answer to Problem 5.48P

The equations have been proven.

Explanation of Solution

The expression for Fourier series is

    f(t)=n=0[ancos(nωt)+bnsin(nωt)]={a0+a1cosωt+a1cosωt+...+ancosnωt+...+b1sinωt+b2sin2ωt+...+bnsinnωt}        (I)

Multiply (I) with cosmωt and integrate within the limit τ2 to τ2 with respect to dt

    f(t)cosmωtdt=τ2τ2a0cos(mωt)dt+τ2τ2a1cosωtcos(mωt)dt+τ2τ2a2cos2ωtcos(mωt)dt+...+τ2τ2ancosmωtcos(mωt)dt+...+τ2τ2b1cosωtcos(mωt)dt+τ2τ2b2cos2ωtcos(mωt)dt+...+τ2τ2bncosnωtcos(mωt)dt+...

But,

    τ2τ2cosnωtcos(mωt)dt={0ifmnτ2ifm=n0

And

  τ2τ2cosnωtcos(mωt)dt=0forallnandm

Therefore all the b terms in the integral will become zero all a terms become zero for except for m=n

Therefore,

    f(t)cosmωtdt=τ2τ2ancosnωtcos(nωt)dt=τ2τ2ancos2nωtdt=τ2τ2an(1+cos(2nωt)2)dt=12τ2τ2andt+τ2τ2(1+cos(2nωt)2)dt        (II)

Here cos(2nωt)2 is an even function from definite integral properties.

Thus τ2τ2(cos(2nωt)2)dt=0

Therefore,

    f(t)cosmωtdt=12τ2τ2andt+0=an2(t)τ2τ2an(τ2)

Rearrange the equation for an

    an=(2τ)f(t)cosmωtdtforn1

For a00 substitute zero for n in (II) and find a0

    f(t)cosmωtdt=τ2τ2a0(1+cos2(0)ωt2)dtτ2τ2f(t)dt=τ2τ2a0(1+12)dt=a0τa0=1ττ2τ2f(t)dt

Now, again multiply (I) with sinmωt and integrate within the limit τ2 to τ2 with respect to dt

f(t)cosmωtdt=τ2τ2a0cos(mωt)dt+τ2τ2a1cosωtsin(mωt)dt+τ2τ2a2cos2ωtsin(mωt)dt+...+τ2τ2ancosmωtsin(mωt)dt+...+τ2τ2b1cosωtsin(mωt)dt+τ2τ2b2cos2ωtsin(mωt)dt+...+τ2τ2bncosnωtsin(mωt)dt+...

But,

    τ2τ2cosnωtsin(mωt)dt={0ifmnτ2ifm=n0

And

  τ2τ2cosnωtsin(mωt)dt=0forallnandm

Therefore all the b terms in the integral will become zero all a terms become zero for except for m=n

Therefore,

    f(t)sinmωtdt=τ2τ2bnsinnωtsin(nωt)dt=τ2τ2bnsin2nωtdt=τ2τ2bn(1cos(2nωt)2)dt=12τ2τ2bndtτ2τ2(cos(2nωt)2)dt

But τ2τ2(cos(2nωt)2)dt=0

Equate and rearrange for bn

    τ2τ2f(t)sinmωtdt=12τ2τ2bndt0=bn2[t]τ2τ2=bnτ2bn=2ττ2τ2f(t)sinmωtdtforn1

And b0=0

Hence proved.

Conclusion:

The equations have been proven.

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