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प्रश्न
Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a period of 2π/ω.
उत्तर
The given equation is in the form of a combination of two harmonic functions. We can write this equation in the form of a single harmonic (sine or cosine) function.
We have displacement function: `y = sin ωt - cos ωt`
`y = sqrt(2) (1/sqrt(2) * sin ωt - 1/sqrt(2) * cos ωt)`
= `sqrt(2)[cos (pi/4) * sin ωt - sin(pi/4) * cos ωt]`
= `sqrt(2) [sin (ωt - pi/4) = sqrt(2) [sin (ωt - pi/4)]]`
⇒ `y = sqrt(2) sin (ωt - pi/4)`
Comparing with the standard equation of S.H.M
`y = a sin(ωt + phi)` we get angular frequency of S.H.M, ω = `(2pi)/T` ⇒ `T = (2pi)/ω`
Hence the function represents S.H.M with a period T = 2π/ω.
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