Analyzing Simple Harmonic Motion

An object hangs on a spring, making it 2 m long in its equilibrium position. See Figure 28. If the object is pulled down 1 m and released, it will oscillate up and down. The length \(l\) of the spring after \( t\) seconds is modeled by the function \(l( t) =2+\cos t\).

1. How does the length of the spring vary?
2. Find the velocity of the object.
3. At what position is the speed of the object a maximum?
4. Find the acceleration of the object.
5. At what position is the acceleration equal to 0?

Solution

1. Since \(l(t) =2+\cos t\) and \(-1 ≤ \cos t ≤ 1,\) the length of the spring oscillates between 1 m and 3 m.
2. The velocity \(v\) of the object is \[ v=l' ( t) =\dfrac{d}{dt}\left( 2+\cos t\right) =-\sin t \]
3. Speed is the magnitude of velocity. Since \(v=-\sin t,\) the speed of the object is \(\left\vert v\right\vert =\left\vert -\sin t\right\vert =\left\vert \sin t\right\vert\). Since \(-1 ≤ \sin t ≤ 1,\) the object moves the fastest when \(\left\vert v\right\vert =\left\vert \sin t\right\vert =1.\) This occurs when \(\sin t=\pm 1\) or, equivalently, when \( \cos t=0.\) So, the speed is a maximum when \(l( t) =2,\) that is, when the spring is at the equilibrium position.
4. The acceleration \(a\) of the object is given by \[ a=l'' ( t) =\dfrac{d}{dt}l' (t)=\dfrac{d}{dt} ( -\sin t) =-\cos t \]
5. Since \(a=-\cos t,\) the acceleration is zero when \(\cos t=0.\) This occurs when \(l( t) =2,\) that is, when the spring is at the equilibrium position. At this time, the speed is maximum.