This is a simple harmonic motion exercise. The proof that the resultant force on the particle, at time t, is -225 Newtons is indicated below.
What is the proof that the resultant force on the particle, at time t, is -225x newtons?
Note that we have been given the following:
Mass = 9kg,
Y = 45 N
ζ = 0.4 m
Then, Y/ζ = 45N/0.4m
= 112.5 N/m
Hence, the spring constant =
K = 112. 5N/m
From this point we can resolve the Amplitude (A):
Amplitude is given as:
A = CB - PB
= 0.6 - 0.5
A = 0.1m
I) So because one spring is elongated by x and the other is compressed by x, Hence, the spring force is:
F[tex]_{net}[/tex] = - Kₓ - Kₓ
= - 2Kₓ
= -2 x 112.5 x* X
F[tex]_{net}[/tex] = -225x Newton ..........Lets call this expression 1
What is the proof that the particles move with simple harmonic motion?
From equation 1 above, we know that
Ma = -225x
a = (-225/m) x
⇒ a ∝ - x,
Hence, the motion is is simple harmonic in nature.
What is the period of the motion?
The time period, T = 2 π√m/K[tex]_{eq}[/tex]
Where, K[tex]_{eq}[/tex] = 2K
= 2π√(9/225)
= 2π * 3/15
= 2(22/7) * (3/15)
= 1.25714285714; thus
T ≈ 1.26 secs...............lets call this expression 2
What is the speed of the particle when it is 0.05 meters from C?
Note that Angular Frequency is = 2π/T
= 2π/1.26
= 4.9866550057
ω = 4.99 rad/s
Since the velocity at any point x from the mean position is given by
V = ω √ (A² - x²); hence
V = 4.99 √[(0.1)² - (0.05)²
V = 4.99 x 0.0866
V = 0.432134
V ≈ 0.432..........Lets call this expression 3
What is the expression of x in terms of t?
Along the x-axis, A simple harmonic motion can be depicted as:
x = A sin (ωt + Ф) ..........lets call this expression 4
where, Ф = Phase constant
At t = 0; x = A
A = A sinФ
⇒ SinФ = 1
⇒ Ф = π/2
Expression 4 becomes:
x = A sin (ωt + π/2)
Inserting the values of ω and A, we have
x = 0.1 sin (4.99t + π/2).........Lets call this expression 5a
or x = 0.1 cos (4.99t).......Lets call this expression 5b
Hence,
Sin(π/2 + Ф) = CosФ
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