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When the displacement of SHO is 1 cm, its velocity is 2 cm/s and when displacement is 2 cm, its velocity is 1 cm/s.Its amplitude A=_____ cm and periodic time T=____s?
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Given Information:
- Displacement at v=2 cm/s: x=1 cm
- Displacement at v=1 cm/s: x=2 cm
Finding Amplitude:
To find the amplitude of the Simple Harmonic Oscillator (SHO), we can use the formula:
A = (x1 - x2) / 2
Substitute the given displacements:
A = (2 - 1) / 2 = 0.5 cm
Therefore, the amplitude of the SHO is 0.5 cm.
Finding Periodic Time:
To find the periodic time of the SHO, we can use the formula:
T = 2π / ω
Where ω is the angular frequency given by:
ω = 2π / T
We know that velocity v is the derivative of displacement x with respect to time t. Using this relationship, we can find the angular frequency ω:
v = dx/dt
Given v = 2 cm/s when x = 1 cm:
2 = dx/dt
Integrating both sides, we get:
2t = x + C
At t = 0, x = 0:
C = 0
Therefore, the displacement x at any time t is given by:
x = 2t
Given v = 1 cm/s when x = 2 cm:
1 = dx/dt
1 = d(2t)/dt = 2
This implies that the maximum displacement occurs at t = 1 s, x = 2 cm.
Substitute in Periodic Time Formula:
Substitute x = 2 cm into x = A*sin(ωt) to find ω:
2 = 0.5*sin(ω)
sin(ω) = 4
This implies that ω = π/2
Finally, substitute ω into the formula for T:
T = 2π / ω = 4π / π = 4 s
Therefore, the amplitude of the SHO is 0.5 cm and the periodic time is 4 s.
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When the displacement of SHO is 1 cm, its velocity is 2 cm/s and when displacement is 2 cm, its velocity is 1 cm/s.Its amplitude A=_____ cm and periodic time T=____s?
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