Respuesta :
Answer:
0.587 m/s
Explanation:
m = mass of the object = 2.7 kg
k = spring constant = 280 N/m
[tex]w[/tex] = angular frequency
Angular frequency is given as
[tex]w =\sqrt{ \frac{k}{m}}[/tex]
[tex]w =\sqrt{ \frac{280}{2.7}}[/tex]
[tex]w[/tex] = 10.2 rad/s
x = position relation to equilibrium position = 0.020 m
A = amplitude
[tex]v[/tex] = speed at position "x" = 0.55 m/s
speed is given as
[tex]v = w\sqrt{A^{2} - x^{2}}[/tex]
[tex]0.55 = (10.2)\sqrt{A^{2} - 0.02^{2}}[/tex]
A = 0.0575 m
[tex]v_{max}[/tex] = maximum speed of the object
maximum speed of the object is given as
[tex]v_{max}=A w[/tex]
[tex]v_{max}=(0.0575) (10.2)[/tex]
[tex]v_{max}[/tex] = 0.587 m/s
The maximum speed attained by the object,
[tex]\rm v_m_a_x=0.587\;m/sec[/tex]
Given :
Mass of the object, m = 2.7 Kg
Spring constant, K = 280 N/m
Speed = 0.55 m/sec
Solution :
We know that the angular velocity is given by,
[tex]\rm \omega = \sqrt{\dfrac{K}{m}}[/tex]
[tex]\rm \omega = \sqrt{\dfrac{280}{2.7}}[/tex]
[tex]\rm \omega = 10.2\;rad/sec[/tex]
Now, speed is given as
[tex]\rm v = \omega\sqrt{A^2-x^2}[/tex]
[tex]0.55=(10.2)\sqrt{A^2-0.02^2}[/tex]
[tex]\rm A = 0.0575\;m[/tex]
Now, maximum speed of the object is,
[tex]\rm v_m_a_x=A\omega[/tex]
[tex]\rm v_m_a_x=0.0575\times10.2=0.587\;m/sec[/tex]
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https://brainly.com/question/12446100?referrer=searchResults
