Irodov 1.99: A Push That Turns Off and Coasts – Amit & Deepali Sinha Foundation

Problem Irodov 1.99

A particle of mass $m$ , at rest at $t=0$ , is set in motion by a force $\vect F=\vect F_0\cos\omega t$ , where $\vect F_0$ is a constant vector and $\omega$ is a constant. Find (a) the time $t$ at which the particle’s speed first returns to zero, (b) the distance $s$ it has covered by that instant, and (c) the maximum speed $v_{\mathrm{max}}$ it reaches at any point during that interval.

Given

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

Find

(a)

t

at the first stop, (b)

s

covered by then, (c)

v_{\mathrm{max}}

over that interval

Solution

Step 1Velocity from Newton’s second law

The force stays along the fixed direction of $\vect F_0$ , so treat the motion as one dimensional along that direction, with $F_0$ and $v$ the scalar force and velocity. Integrating from rest at $t=0$ ,

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

v(t) = \dfrac{F_0}{m\omega}\sin\omega t

(1)

Step 2The first instant the particle stops

By (1), $v(t)=0$ whenever $\sin\omega t=0$ , at $\omega t=0,\pi,2\pi,\dots$ . Between the start and the first later zero, at $\omega t=\pi$ , $\sin\omega t>0$ , so the particle genuinely moves forward and then comes back to rest there for the first time.

t = \dfrac{\pi}{\omega}

(2)

Step 3Distance covered by the first stop

Integrating (1) from $0$ to the general time $t$ , and since $v(t)\ge0$ throughout $[0,\pi/\omega]$ (Step 2), this is the actual distance traveled:

\begin{align*}s(t) &= \frac{F_0}{m\omega}\int_0^t \sin\omega t'\,dt' \\ &= \frac{F_0}{m\omega}\left.\left(-\frac{\cos\omega t'}{\omega}\right)\right|_0^t \\ &= \frac{F_0}{m\omega^2}\bigl(1-\cos\omega t\bigr)\end{align*}

Evaluating at the stopping time $t=\pi/\omega$ found in Step 2, where $\cos\pi=-1$ ,

s = \frac{F_0}{m\omega^2}\bigl(1-(-1)\bigr).

s = \dfrac{2F_0}{m\omega^2}

(3)

Step 4The maximum speed over the interval

The speed , in units where F_0/(mω)=1, plotted against ω t. It rises from rest, peaks at ω t=π/2, where the driving force passes through zero, and returns to zero at ω t=π. — The speed (1), in units where $F_0/(m\omega)=1$ , plotted against $\omega t$ . It rises from rest, peaks at $\omega t=\pi/2$ , where the driving force passes through zero, and returns to zero at $\omega t=\pi$ .

On $[0,\pi/\omega]$ , (1) is $F_0/(m\omega)$ times $\sin\omega t$ , which is largest at $\omega t=\pi/2$ , the interval’s midpoint (marked on the plot), where $\sin(\pi/2)=1$ .

v_{\mathrm{max}} = \dfrac{F_0}{m\omega}

(4)

Check

Dimensionally, $[F_0]/([m][\omega])=\mathrm N/(\mathrm{kg\cdot s^{-1}})=\mathrm{m/s}$ , a speed, and $[F_0]/([m][\omega]^2)$ is a length, so both boxed results in (b) and (c) check. The average speed over the interval is $s/t=2F_0/(\pi m\omega)$ , and $v_{\mathrm{max}}/(s/t)=\pi/2\approx1.57$ , exactly the known ratio of a sine arch’s peak to its own mean value.

Remark

Unlike the companion force $F_0\sin\omega t$ (problem 1.98), whose velocity $v=(F_0/m\omega)(1-\cos\omega t)$ never turns negative, this cosine-driven velocity $v=(F_0/m\omega)\sin\omega t$ genuinely changes sign at $\omega t=\pi$ : past that instant the particle does not merely pause, it reverses and retraces its path. The two forces differ only by a phase, but that phase decides whether the particle’s rest points are momentary pauses or true turning points.