A body of 0.25 kg executes SHM given by y = 0.4 sin 0.5mt m. Calculate the amplitude, angular frequency, maximum velocity and maximum acceleration.
Answers
Given info : A body of 0.25 kg executes SHM given by y = 0.4 sin 0.5πt m.
To find :
- amplitude
- angular frequency
- maximum velocity
- maximum acceleration
solution : equation of simple harmonic motion is y = 0.4 sin 0.5πt ...(1)
we know, standard equation of simple harmonic motion is y = Asinωt ...(2)
where A is amplitude of SHM, ω is angular frequency.
on comparing we get,
amplitude = A = 0.4 m
angular frequency = ω = 0.5π rad/s
maximum velocity of particle in SHM is given by, v = ωA
so, v = 0.5π × 0.4 = 0.2π m/s = 0.628 m/s²
maximum acceleration of particle in SHM is given by, a = ω²A
= (0.5π)² × 0.4 = 0.25π² × 0.4 = 0.98596 m/s²
Answer:
Given info : A body of 0.25 kg executes SHM given by y = 0.4 sin 0.5πt m.
To find :
amplitude
angular frequency
maximum velocity
maximum acceleration
solution : equation of simple harmonic motion is y = 0.4 sin 0.5πt ...(1)
we know, standard equation of simple harmonic motion is y = Asinωt ...(2)
where A is amplitude of SHM, ω is angular frequency.
on comparing we get,
amplitude = A = 0.4 m
angular frequency = ω = 0.5π rad/s
maximum velocity of particle in SHM is given by, v = ωA
so, v = 0.5π × 0.4 = 0.2π m/s = 0.628 m/s²
maximum acceleration of particle in SHM is given by, a = ω²A
= (0.5π)² × 0.4 = 0.25π² × 0.4 = 0.98596 m/s²
Explanation: