First slide

Simple harmonic motion

Question

An object executes periodic motion which is given by $\sin 4 π t - c o s 4 π t .$ Then we can say that

Moderate

A

It executes simple harmonic motion with a time period of T = 2s
B

It is periodic but not simple harmonic with a time period of T = 2s
C

It executes simple harmonic motion with a time period of T = ½ s
D

It is periodic but not simple harmonic with a time period of T = ½ s

Solution

The function $\sin 4 π t - c o s 4 π t$ will represent a periodic motion, if it is identically repeated after a fixed interval of time. If this is also a simple harmonic motion. Then we can uniquely write that as $A \sin ω t + ϕ$
$∴ \sin 4 π t - c a s 4 π t$ can be reduced as follows
$\sqrt{2} (\frac{1}{\sqrt{2}} \sin 4 π t - \frac{1}{\sqrt{2}} c o s 4 π t)$ (multiplying & dividing by $\sqrt{2}$ )
$= \sqrt{2} [\cos (\frac{π}{4}) \sin (4 π t) - \sin \frac{π}{4} c o s (4 π t)] (∵ \sin \frac{π}{4} = \cos \frac{π}{4} = \frac{1}{\sqrt{2}})$
The above relation can be written as
$= \sqrt{2} [\sin π t - \frac{π}{4}] (∵ \sin A \cos B - \cos A \sin B = \sin (A - B))$
The above equation can be compared with $y = A \sin (ω t + ϕ)$ This show that the function $\sin (4 π t - c o s 4 π t)$ is simple harmonic with its resultant as $\sqrt{2} [\sin π t - \frac{π}{4}]$ . Its time period can be found using $ω = \frac{2 π}{T}$
$\begin{array}{l} Here ω = 4 π, \\ ω = 4 π \\ \frac{2 π}{T} = 4 π \\ T = \frac{1}{2} s \end{array}$

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