Question

Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) non-periodic motion? Give period for each case of periodic motion (ω is any positive constant):

(a) sin ωt – cos ωt

(b) sin3 ωt

(c) 3 cos (π/4 – 2ωt)

(d) cos ωt + cos 3ωt + cos 5ωt

(e) exp (–ω2t2)

(f) 1 + ωt + ω2t2

Answer 1

(a) sinwt - coswt .
= √2 { 1/√2 sinwt - 1/√2coswt}
= √2{cos45°.sinwt - sin45°.coswt}
= √2sin(wt - 45°)
This is in the form of y = Asin(wt ± ∅)
So, this is the equation of SHM .
Period = 2π/w

(b) sin³wt
Use formula of sin3x = 3sinx - 4sin³x
So, sin³wt = 1/4 [ 3sinwt - sin3wt ]
Hence, you observed that this equation is combination of two SHM. Hence, this is not SHM. But periodic motion .
Period = LCM of period { sinwt , sin3wt }
Period of sinwt = 2π/w
period of sin3wt = 2π/3w
so, period = 2π/w

(c) 3cos(π/4 - 2wt)
= 3cos {-(2wt-π/4)}
= 3cos(2wt - π/4) [ cos(-∅) = cos∅]
Hence, this is SHM .
Period = 2π/2w = π/w


(d) coswt + cos3wt + cos5wt
This isn't SHM. But perioic motion.
Period = LCM of period { coswt, cos3wt,cos5wt}
= LCM of { 2π/w, 2π/w ,2π/5w}
= 2π/w

(e) x = exp( -w²t²)
x = e^(-w²t²) this is an exponential function it decreases monotonically .
Also at x →0 t →∞
It means there is no repetation .
So, this is neither Periodic nor SHM

(f) x = 1 + wt + w²t²
At x →∞ , t→∞
There is no repetation of values . hence, this neither periodic nor SHM.