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Q.A simple harmonic oscillator is represented by the equation : Y = 0.40 sin (440t + 0.61) where Y is in metres and t is in seconds. Find the values of
  1. Frequency of oscillation
  2. Time period of oscillation and
  3. Initial phase

Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2022Subjective· 3mImportance★★★★★
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Matching Y=0.40sin⁡(440t+0.61)Y = 0.40\sin(440t + 0.61) against the standard SHM equation y=Asin⁡(ωt+ϕ)y = A\sin(\omega t + \phi) directly gives ω=440 rad/s\omega = 440\ \text{rad/s} and ϕ=0.61 rad\phi = 0.61\ \text{rad}, from which frequency and time period follow.

The general equation of simple harmonic motion is:

y=Asin⁡(ωt+ϕ)y = A\sin(\omega t + \phi)

where AA is the amplitude, ω\omega is the angular frequency, and ϕ\phi is the initial phase (phase constant).

Comparing with the given equation Y=0.40sin⁡(440t+0.61)Y = 0.40\sin(440t + 0.61):

  • Amplitude: A=0.40 mA = 0.40\ \text{m}
  • Angular frequency: ω=440 rad/s\omega = 440\ \text{rad/s}
  • Initial phase: ϕ=0.61 rad\phi = 0.61\ \text{rad}

1. Frequency of oscillation:

f=ω2π=4402π≈4406.283≈70.03 Hzf = \dfrac{\omega}{2\pi} = \dfrac{440}{2\pi} \approx \dfrac{440}{6.283} \approx 70.03\ \text{Hz}

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