Irodov 1.98: Distance Under a Sinusoidal Push – Amit & Deepali Sinha Foundation

Problem Irodov 1.98

A particle of mass $m$ , at rest at $t=0$ , is set in motion by a force $\vect F = \vect F_0\sin\omega t$ , where $\vect F_0$ is a constant vector and $\omega$ is a constant. Find the distance $s$ the particle has covered as a function of $t$ , and sketch $s(t)$ .

Given

$m$
$\vect F_0$
$\omega$
particle at rest at $t=0$

Find

s(t)

, and a sketch of that function

Solution

Step 1Velocity from Newton’s second law

The force stays along the fixed direction of $\vect F_0$ , so the motion is one dimensional along that direction; write $F_0$ and $v$ for the scalar force and velocity along it. Newton’s second law integrates from rest at $t=0$ to a general time $t$ :

\int_0^v dv' = \frac{F_0}{m}\int_0^t \sin\omega t'\,dt'.

\begin{aligned} v(t) &= \frac{F_0}{m}\left.\left(-\frac{\cos\omega t'}{\omega}\right)\right|_0^t \\ &= \frac{F_0}{m\omega}\bigl(1-\cos\omega t\bigr) \end{aligned} \tag{1}

Step 2Distance from the velocity

The distance covered, in units where F_0/(mω^2)=1, plotted against ω t. The curve never decreases; watch the plateau it flattens into once every period, at ω t = 2π n, where the speed momentarily returns to zero. — The distance covered, in units where $F_0/(m\omega^2)=1$ , plotted against $\omega t$ . The curve never decreases; watch the plateau it flattens into once every period, at $\omega t = 2\pi n$ , where the speed momentarily returns to zero.

Integrating (1) from $0$ to $t$ gives the distance covered, and because $v(t)$ never goes negative (Step 1), this integral is the actual path length, not just the net displacement.

\begin{align*}s(t) &= \int_0^t v(t')\,dt' \\ &= \frac{F_0}{m\omega}\int_0^t\bigl(1-\cos\omega t'\bigr)\,dt' \\ &= \frac{F_0}{m\omega}\left.\left(t'-\frac{\sin\omega t'}{\omega}\right)\right|_0^t\end{align*}

s(t) = \dfrac{F_0}{m\omega^2}\bigl(\omega t-\sin\omega t\bigr)

(2)

Check

Dimensionally, $[F_0]/[m][\omega]^2 = \mathrm N/(\mathrm{kg\cdot s^{-2}}) = \mathrm{kg\cdot m/s^2}\cdot\mathrm{s^2/kg}=\mathrm m$ , a length, and $\omega t-\sin\omega t$ is dimensionless, as required. For small $\omega t$ , expanding $\sin\omega t\approx\omega t-(\omega t)^3/6$ gives $s\approx F_0\omega t^3/(6m)$ , matching the elementary result for a force ramped up as $F\approx F_0\omega t$ from rest.

Remark

The speed $v(t)=(F_0/m\omega)(1-\cos\omega t)$ returns to exactly zero once every period, at $\omega t=2\pi n$ , which is what produces the flat plateaus on the curve above, but it never turns negative, so the particle only pauses there and never reverses. A force oscillating symmetrically about zero can still push a particle steadily one way, because the direction of travel is set by the sign of an integral of the force, not by the sign of the force at any one instant.