Respuesta :
Answer:
t_1 = 0.5*pi*sqrt( m / k )
Explanation:
Given:
- The block of mass m undergoes simple harmonic motion. With the displacement of x from mean position is given by:
x(t) = A*cos(w*t)
Find:
- At what time t1 does the block come back to its original equilibrium position (x=0) for the first time?
Solution:
- The first time the block moves from maximum position to its mean position constitutes of 1/4 th of one complete cycle. So, the required time t_1 is:
t_1 = 0.25*T
- Where, T : Time period of SHM.
- The time period for SHM is given by:
T = 2*pi*sqrt ( m / k )
Hence,
t_1 = 0.25 * 2 * pi * sqrt( m / k )
t_1 = 0.5*pi * sqrt( m / k )
The time that it will take the block to get back to its' original equilibrium position for the first time is; t₁ = 0.5π√(m/k)
What is the time to reach the equilibrium position?
From general formula of simple harmonic motion, the mean position is given by:
x(t) = A*cos(ω*t)
The first time the block moves from maximum position to its mean position will be ¹/₄ th of one complete cycle. Thus, the required time t₁ is:
t₁ = 0.25*T
Where;
T is time period of SHM with the formula;
T = 2π√(m/k)
Thus;
t₁ = 0.25 * 2 * π * √(m/k)
t₁ = 0.5π√(m/k)
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