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An undamped 2.89-kg horizontal spring oscillator has a spring constant of 24.3 n/m. while oscillating, it is found to have a speed of 3.89 m/s as it passes through its equilibrium position. what is its amplitude of oscillation?

Respuesta :

skyluke89
The speed of the system when it passes through the equilibrium position is [tex]v=3.89 m/s[/tex], and this corresponds to the maximum speed of the oscillator.

The angular frequency of the oscillator is given by
[tex]\omega= \sqrt{ \frac{k}{m} } [/tex]
where k is the spring constant, and m the mass. Substituting the data, we find
[tex]\omega= \sqrt{ \frac{24.3 N/m}{2.89 kg} }=2.9 s^{-1} [/tex]

and since the amplitude of the oscillation, A, is related to the maximum speed by
[tex]v_{max}=\omega A[/tex]
we can re-arrange this equation to calculate A:
[tex]A= \frac{v_{max}}{\omega}= \frac{3.89 m/s}{2.9 s^{-1}}=1.34 m [/tex]
nuhulawal20

As the undamped spring oscillator passes through its equilibrium position, its amplitude of oscillation is 1.34m.

Given the data in the question;

  • Mass of the undamped; [tex]m = 2.89 kg[/tex]
  • Spring constant; [tex]k = 24.3 n/m[/tex]
  • Speed; [tex]v = 3.89 m/s[/tex]

Amplitude of oscillation; [tex]A =\ ?[/tex]

Using Conservation of Energy:

[tex]\frac{1}{2}mv^2 = \frac{1}{2}kA^2[/tex]

Where, m is the mass, v is the speed, k is the spring constant and A is the amplitude of oscillation

We substitute our given values into the equation

[tex]\frac{1}{2}*2.89kg*(3.89m/s)^2 = \frac{1}{2}* 24.3N/m * A^2\\\\21.86588kg.m^2/s^2 = 12.15N/m * A^2\\\\21.86588N.m = 12.15N/m * A^2\\\\A^2 = \frac{21.86588N.m}{12.15N/m} \\\\A^2 = 1.79966 m^2\\\\A = \sqrt{1.79966 m^2}\\\\A = 1.34m[/tex]

Therefore, as the undamped spring oscillator passes through its equilibrium position, its amplitude of oscillation is 1.34m.

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