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Show that the motion of a particle represented by y = sin ωt – cos ωt is simple harmonic with a period of 2π/ω.

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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.

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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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पाठ 14: Oscillations - Exercises [पृष्ठ १०३]

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एनसीईआरटी एक्झांप्लर Physics Exemplar [English] Class 11
पाठ 14 Oscillations
Exercises | Q 14.30 | पृष्ठ १०३

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