Question
The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds) as $x=0.05 \cos \left(4 \pi t+\frac{\pi}{4}\right)$. The frequency of the motion will be [MP PMT / PET 1998](a) $\mathrm{O} .5 \mathrm{~Hz}$(b) $1.0 \mathrm{~Hz}$(c) $1.5 \mathrm{~Hz}$(d) $2.0 \mathrm{~Hz}$
Step 1
Step 1: The standard equation of simple harmonic motion is given by $y = A \sin(\omega t + \phi)$, where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is the time, and $\phi$ is the phase constant. Show more…
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The displacement $x$ (in metre) of a particle performing simple harmonic motion is related to time $t$ (in second) as $x=0.05 \cos \left(4 \pi t+\frac{\pi}{4}\right)$. The frequency of the motion will be (a) $0.5 \mathrm{~Hz}$ (b) $1.0 \mathrm{~Hz}$ (c) $1.5 \mathrm{~Hz}$ (d) $2.0 \mathrm{~Hz}$
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