Answer:
X(t) = 13/13 cos(12t+α)
C =13/13
π/6 s
Explanation:
(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.
(a) Find X(t) in the form c • cos(w_o*t— α)
(b) Find the amplitude 3 Period of motion of the body 1
mass: m = 200g = 0.200 kg
displacement: ΔX = 20 cm = 0.20 m
Spring Constant: K = 9/0.20 = 45 N/m
IV: X(0) = 1m V(0) = -5 m/s
Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)
Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s
X(0) = 1m =c_1
X'(0) = V(0) = c_2*w_o/w_o
= -5/12 = c_2
"radians Technically Unitless"
Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c
X(t) = 13/13 cos(12t+α)
since, C>0 : damped forced vibration c_1>0, c_2>0
phase angle 2π+tan^-1(c_2/c_1)
=2π+tan^-1(-5/12/1)= 5.884
period: T =2π/w_o
=π/6 s
(a) The position as a function of time is given by: x(t) = cos(15t + α)
(b) The time period is 0.418s
The mass of the body is, m = 200g = 0.2kg
The spring is stretched by a distance, x = 20cm = 0.2m
By a force F = 9N
The restoring force of the spring is given by:
F = kx
where k is the spring constant
9 = k × (0.2)
k = 45 N/m
Now the angular frequency of the body in SHM is given by:
ω = [tex]\sqrt{\frac{k}{m} }[/tex]
ω = [tex]\sqrt{\frac{45}{0.2} }[/tex]
ω = 15 rad/s
The amplitude of the oscillation will be C = 1 since the spring is stretched by 1 m.
The position of the body as a function of time is:
x(t) = Ccos(ωt + α)
x(t) = cos(15t + α)
The time period of oscillation is defined as:
[tex]T=\frac{2\pi}{\omega}[/tex]
T = 2π/15
T = 0.418s
Learn more about simple harmonic motion:
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