Ncert Exercises & Solutions

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

Hint: It should be a periodic function comparable to the standard equation of motion.

Step 1: Find the equation of a single SHM.

We have to convert the given combination of two harmonic functions to a single harmonic (sine or cosine) function.

Given, displacement function, y= sin

ω t

- cos

ω t

= \sqrt{2} (\frac{1}{\sqrt{2}} \cdot \sin ω t - \frac{1}{\sqrt{2}} \cdot \cos ω t)

= \sqrt{2} [\cos (\frac{π}{4}) \cdot \sin ω t - \sin (\frac{π}{4}) \cdot \cos ω t]

= \sqrt{2} [\sin (ω t - \frac{π}{4})]

Step 2: Find the time period of the oscillation.

Comparing it with the standard equation, y=asin

(ω t + ϕ)

we get,

ω = \frac{2 π}{T} \Rightarrow T = \frac{2 π}{ω}

Clearly, the function represents SHM with a period

T = \frac{2 π}{ω}