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Displacement of a particle at any instant of time t is y = 5 sin(100\(\pi\)t + \(\phi\)). The frequency of oscillation of the particle is:
Displacement of a particle at any instant of time t is y = 5 sin(100\(\pi\)t + \(\phi\)). The frequency of oscillation of the particle is:
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Always be careful to distinguish between angular frequency \(\omega\) (in rad/s) and frequency \(f\) (in Hz). The term multiplying \(t\) inside the sine/cosine function is always \(\omega\).
Updated On: Feb 10, 2026
- 100 Hz
- 25 Hz
- 200 Hz
- 50 Hz
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The Correct Option is D
Solution and Explanation
Step 1: State the standard equation for simple harmonic motion (SHM).
The displacement \(y\) in SHM is defined by \(y = A \sin(\omega t + \phi)\), where \(A\) represents amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase angle.
Step 2: Isolate the angular frequency \(\omega\) from the provided equation.
The given equation is \(y = 5 \sin(100\pi t + \phi)\). By comparing this to the standard form, it is evident that the angular frequency \(\omega = 100\pi\) rad/s.
Step 3: Convert angular frequency (\(\omega\)) to linear frequency (\(f\)).
The governing relationship is \(\omega = 2\pi f\).
\[ f = \frac{\omega}{2\pi} \]
Step 4: Compute the frequency.
\[ f = \frac{100\pi}{2\pi} = 50 \, \text{Hz} \]
The oscillation frequency is 50 Hz.
The displacement \(y\) in SHM is defined by \(y = A \sin(\omega t + \phi)\), where \(A\) represents amplitude, \(\omega\) is angular frequency, and \(\phi\) is the phase angle.
Step 2: Isolate the angular frequency \(\omega\) from the provided equation.
The given equation is \(y = 5 \sin(100\pi t + \phi)\). By comparing this to the standard form, it is evident that the angular frequency \(\omega = 100\pi\) rad/s.
Step 3: Convert angular frequency (\(\omega\)) to linear frequency (\(f\)).
The governing relationship is \(\omega = 2\pi f\).
\[ f = \frac{\omega}{2\pi} \]
Step 4: Compute the frequency.
\[ f = \frac{100\pi}{2\pi} = 50 \, \text{Hz} \]
The oscillation frequency is 50 Hz.
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