Answer :
If its equilibrium position and given a downward velocity of 1.2 m/s, determine the position of the block when t =0.22 s then the position of the block when t = 0.22 s is 0.046 m.
How to find?
Using the equation of motion for simple harmonic motion, we can determine the position of the block as follows:
x = [tex]Acos(ωt + ϕ) + x₀[/tex]
Where;
x = position of the block
A = amplitude
ω = angular frequency
t = time elapsedϕ
= phase constant
x₀ = initial displacement
Also, since it is displaced 91 mm downward from its equilibrium position, its initial displacement is
x₀ = -91 mm
= -0.091 m.
Hence, substituting all these values in the formula above;
x = [tex]Acos(ωt + ϕ) + x₀[/tex]
We have;
x = [tex]-0.091 cos(2.093t + ϕ) + 0.091.[/tex]
Taking the derivative of the position function, we can obtain the velocity function for the block as follows;
v = [tex]-Aω sin(ωt + ϕ).[/tex]
Therefore, we need to find the phase constant ϕ such that the displacement x and velocity v of the block at time t = 0.22 s is as follows;
x = -0.091 cos(2.093t + ϕ) + 0.091
= -0.091 cos(2.093 × 0.22 + ϕ) + 0.091
= -0.0576 + 0.091 cos ϕv
= 0.191 sin(2.093t + ϕ)
= 0.191 sin(2.093 × 0.22 + ϕ)
= 0.128 sin ϕAt time t = 0.22 s,
the velocity of the block is given as 1.2 m/s, and it is moving downwards.
Therefore;
1.2 = 0.191 sin ϕ.
Since sin ϕ can take only values between -1 and 1, the possible values of ϕ are as follows;
ϕ₁ = sin⁻¹(1.2/0.191)
= 1.270 radϕ₂
= π - sin⁻¹(1.2/0.191)
= 1.871 rad.
We can then find the displacement of the block when t = 0.22 s as follows;
x = -0.091 cos(2.093t + ϕ) + 0.091
= -0.091 cos(2.093 × 0.22 + 1.871) + 0.091
= 0.046 m.
Therefore, the position of the block when t = 0.22 s is 0.046 m.
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