This question involves the concepts of simple harmonic motion and maximum speed.
The maximum speed of the child is "0.83 m/s".
The maximum speed in the simple harmonic motion is given by the following formula:
[tex]v=A\omega[/tex]
where,
v = maximum speed = ?
A = Amplitude = 0.29 m
T = period = 2.2 s
ω = angular frequency = [tex]\frac{2\pi}{T} = \frac{2\pi}{2.2\ s} = 2.86\ rad/s[/tex]
Therefore,
v = (0.29 m)(2.86 rad/s)
v = 0.83 m/s
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The expressions of the simple harmonic movement allows to find the maximum speed of the swing is:
The simple harmonic movement is a periodic movement where the restoring force is proportional to the elongation, in the case of a swing it can be approximated to a simple pendulum which is a mass with an inextensible chord., For small oscillations (θ <15º), is described by the expression.
θ = θ₀ cos (wt + Ф)
w² = L / g
Where θ are the angles, θ₀ the initial angle, w the angular velocity, t the time and Ф a phase constant that is determined by the initial conditions, g the acceleration of gravity and L the length of the pendulum.
Speed is defined in kinematics.
w =[tex]\frac{d \theta}{dt}\\[/tex]
w = [tex]- \theta_o w \ \frac{d \theta }{dt}[/tex]
The speed is maximum when the sine function is equal to ±1
w = [tex]\theta_o w[/tex]
The angular velocity is related to the period.
w = [tex]\frac{2\pi }{T}[/tex]
Let's replace.
w = [tex]\theta_o \ \frac{2 \pi }{T}[/tex]
In rotational motion the eels must be in radians y and the linear and angular variables are related.
[tex]\theta = s/R\\s = R\theta[/tex]
v = w R
w = [tex]\frac{v}{R}[/tex]
We substitute.
v = [tex]\frac{2 \pi \ s}{T}[/tex]
Let's calculate
v = [tex]2 \pi \frac{0.290}{2.20}[/tex]2pi 0.290 / 2.20
v = 0.828 m / s
In conclusion using the expressions of the simple harmonic motion we can find the maximum speed of the swing is:
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