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A particle of mass 2 kg is executing S.H.M. given by y = 6.0 cos (100 + π/4) cm. Its maximum kinetic energy is ________ J.
Correct answer is '36'. Can you explain this answer?
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A particle of mass 2 kg is executing S.H.M. given by y = 6.0 cos (100 ...
Vmax = 600 cm/sec = 6m/sec
K.E. = 36 J
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A particle of mass 2 kg is executing S.H.M. given by y = 6.0 cos (100 ...
T) cm, where t is in seconds. Find the amplitude, time period, frequency, and maximum velocity of the particle.
Amplitude: The amplitude of the SHM is given by the coefficient of cos in the equation. In this case, the amplitude is 6.0 cm.
Time period: The time period of the SHM is given by T = 2π/ω, where ω is the angular frequency. In this case, the angular frequency is 100 rad/s (since it is the coefficient of t in the equation), so the time period is T = 2π/100 = 0.063 seconds.
Frequency: The frequency of the SHM is given by f = 1/T. In this case, the frequency is f = 1/0.063 = 15.87 Hz.
Maximum velocity: The maximum velocity of the particle occurs at the equilibrium position (when y = 0), and is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. In this case, vmax = 6.0 cm x 100 rad/s = 600 cm/s.
Amplitude: The amplitude of the SHM is given by the coefficient of cos in the equation. In this case, the amplitude is 6.0 cm.
Time period: The time period of the SHM is given by T = 2π/ω, where ω is the angular frequency. In this case, the angular frequency is 100 rad/s (since it is the coefficient of t in the equation), so the time period is T = 2π/100 = 0.063 seconds.
Frequency: The frequency of the SHM is given by f = 1/T. In this case, the frequency is f = 1/0.063 = 15.87 Hz.
Maximum velocity: The maximum velocity of the particle occurs at the equilibrium position (when y = 0), and is given by vmax = Aω, where A is the amplitude and ω is the angular frequency. In this case, vmax = 6.0 cm x 100 rad/s = 600 cm/s.
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