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A \(50.0-\mathrm{g}\) object is attached to a horizontal spring with a force constant of \(10.0 \mathrm{~N} / \mathrm{m}\) and released from rest with an amplitude of \(25.0 \mathrm{~cm}\). What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?

Short Answer

Expert verified
The velocity of the object when it is halfway to the equilibrium position is \(1.03 \, m/s\).

Step by step solution

01

Defining given values and the equation for simple harmonic motion

The given values are:- the mass \(m = 50.0 \, g = 0.05 \, kg\) (converted to kg as SI unit for mass is kg),- the spring constant \(k = 10.0 \, N/m\), - and the amplitude \(A = 25.0 \, cm = 0.25 \, m\).The object is released from rest at the amplitude A, meaning initial velocity \(v_0 = 0 \, m/s\), and it's asked for the velocity when the object is halfway to the equilibrium, i.e., at \(x = A/2 = 0.125 \, m\).The equation for simple harmonic motion is: \[v = \sqrt{\frac{k}{m}} \sqrt{A^2 − x^2}\]where \(v\) is the instantaneous velocity of the object; \(k\) the spring constant; \(m\) the mass; \(A\) the amplitude (maximum displacement) and \(x\) the displacement.
02

Substitution of values into the equation

Substitute the given values into the equation:\[v = \sqrt{\frac{10.0 \, N/m}{0.05 \, kg}} \sqrt{(0.25 \, m)^2 − (0.125 \, m)^2}\]
03

Calculation

Now, compute the value to gain the velocity:\[v = \sqrt{200 \, s^–2} \sqrt{0.25^2 \, m^2 − 0.125^2 \, m^2} = 14.14 \, m/s \sqrt{0.046875 \, m^2} = 1.03 \, m/s\]

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Most popular questions from this chapter

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