A force F = Asinω t (A , ω constants) starts acting on a…

A force $F = A\sin\omega t$ ($A$, $\omega$ constants) starts acting on a body of mass $m$ at rest. (a) Find the velocity as a function of time. (b) Find the distance travelled in time $t$.

Explanation of Method

Motion under a sinusoidal force (second version).

Step 1: (a) Velocity

\[ m\frac{dv}{dt} = A\sin\omega t, \]

\[ v(0) = 0. \]

\[ v(t) = \frac{A}{m}\int_0^t \sin\omega t'\,dt' = \frac{A}{m\omega}\big[-\cos\omega t'\big]_0^t. \]

\[v(t) = \frac{A}{m\omega}\,(1 - \cos\omega t)\]

Step 2: (b) Distance

Since $v \ge 0$ at all times, the body never moves backward, and the distance equals the displacement: $s(t) = \int_0^t v\,dt' = \frac{A}{m\omega}\left[t' - \frac{\sin\omega t'}{\omega}\right]_0^t$.

\[s(t) = \frac{A}{m\omega}\left(t - \frac{\sin\omega t}{\omega}\right)\]

$\) \(\omega t$ $-1$ $1$ $2$ $\pi$ $2\pi$ $3\pi$ $4\pi$ $F/A$ $v \,/\, \tfrac{A}{m\omega}$

The force (dashed) alternates in direction, but the velocity $\tfrac{A}{m\omega}(1 - \cos\omega t)$ never becomes negative: the body only speeds up and slows down, and stops for an instant at $\omega t = 2\pi, 4\pi, \dots$

Result

\[\boxed{\,v(t) = \frac{A}{m\omega}\,(1 - \cos\omega t)\,}\]

\[\boxed{\,s(t) = \frac{A}{m\omega}\left(t - \frac{\sin\omega t}{\omega}\right)\,}\]

Remark

Note. This problem is the same as Translational Dynamics 059; the original results are correct. Although the force keeps reversing, the velocity never does: during each half-period with a backward force, the body only loses the speed it gained in the previous half-period.