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Q13.4 (a):
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): (\sin \omega t - \cos \omega t)

Solution :

Step 1: Analyzing the Function Form

The given function is $f(t) = \sin(\omega t) - \cos(\omega t)$. To determine if this represents Simple Harmonic Motion (SHM), we must transform it into a single sinusoidal function of the form $A \sin(\omega t + \phi)$ or $A \cos(\omega t + \phi)$.

Step 2: Algebraic Transformation

Multiply and divide the expression by $\sqrt{2}$ to utilize trigonometric identities:

$f(t) = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin(\omega t) - \frac{1}{\sqrt{2}} \cos(\omega t) \right)$

[By the Trigonometric Identity: $\sin(A - B) = \sin A \cos B - \cos A \sin B$]

Substitute $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ and $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$:

$f(t) = \sqrt{2} \left( \sin(\omega t) \cos(\frac{\pi}{4}) - \cos(\omega t) \sin(\frac{\pi}{4}) \right)$

$f(t) = \sqrt{2} \sin\left(\omega t - \frac{\pi}{4}\right)$

Step 3: Verification of Simple Harmonic Motion

The standard definition of Simple Harmonic Motion requires that the displacement $x(t)$ can be expressed as a linear combination of $\sin(\omega t)$ and $\cos(\omega t)$ [Per the definition of SHM in linear dynamics].

Since the resulting function $f(t) = \sqrt{2} \sin(\omega t - \frac{\pi}{4})$ adheres to the form $A \sin(\omega t + \phi)$, where amplitude $A = \sqrt{2}$ and phase constant $\phi = -\frac{\pi}{4}$, the motion is strictly Simple Harmonic.

Step 4: Determining the Period

For any function of the form $\sin(\omega t + \phi)$, the angular frequency is $\omega$. The time period $T$ is defined by the relation:

$T = \frac{2\pi}{\omega}$

[Per the definition of the time period of a harmonic oscillator]

Visual Representation of the Harmonic Waveform

t f(t) f(t) = √2 sin(ωt - π/4)
Property Value/Status
Motion Type Simple Harmonic Motion
Amplitude $\sqrt{2}$
Angular Frequency $\omega$
Time Period $\frac{2\pi}{\omega}$

Final Solution: The function $(\sin \omega t - \cos \omega t)$ represents Simple Harmonic Motion with a time period $T = \frac{2\pi}{\omega}$.


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