Question

A 175175-g\mathrm{g} glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m155 \mathrm{~N} / \mathrm{m}. At the instant you make measurements on the glider, it is moving at 0.8150.815 m//s and is 3.003.00 cm from its equilibrium point. Use energy\it{energy} conservation\it{conservation} to find
(cc) What is the angular frequency of the oscillations?

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Answered 5 months ago
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(c) Using the expressions of amaxa_{max} and ww we could get the relationship between the frequency and the spring constant from Newton's law of motion by

F=mamkxm=m(ω2xm)ω=km\begin{align*} F = m a_{m}\\ k x_m = m (\omega^{2} x_{m} )\\ \omega = \sqrt{\dfrac{k}{ m}} \tag{5} \end{align*}

Now we can plug our values for mm and kk into equation (5) to get ω\omega

ω=km=155 N/m0.175 kg=29.8 rad/s\begin{align*} \omega &= \sqrt{\dfrac{k}{ m}} \\ &= \sqrt{\dfrac{155\mathrm{~N/m} }{ 0.175 \mathrm{~kg} }}\\ & = \boxed{29.8 \mathrm{~rad/s} } \end{align*}

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