Question

Graham, a physics enthusiast, is asked by his instructor to visit a nearby carpentry. He observed
there a hand–held powered reciprocating saw, say a sabre saw, which performs SHM in a vertical plane with
an angular frequency of 350 rad s−1
. Assume that the tip of the toothed blade of the saw is at x = 0 with
positive velocity when t = 0 and the amplitude of SHM is 2.40 cm. Write an equation, as a function of time,
which shows the position of the tip of the toothed blade. Also calculate that how much will the blade take to
travel from x = 0 to x = 1.20 cm? To 2.40 cm?

Answer 1

Let the position of the tip of the toothed blade be represented by the SHM equation $x=A\ sin(\omega t+\phi)$ ............(1)

where $x$ is the displacement of the tip at time $t$ , $A$ is the amplitude, $\omega$ is the angular frequency, and $\phi$ is the initial phase.

Given $\omega=350\ rad/s$ , $A=2.40\ cm$

Is it also given that at $t=0,$ $x=0$

Therefore, from (1),

$0=A\ sin(0+\phi)\\ \Rightarrow sin\phi=0\\ \Rightarrow \phi=0$

$\therefore$ From (1), $x=A\ sin(\omega t+0)$

$\Rightarrow x=A\ sin(\omega t)$

Velocity, $v=\frac{dx}{dt}=A\omega\ cos(\omega t)$

At $t=0$ , $v=A\omega$ , which is positive and consistent with the problem.

So, $x=A\ sin(\omega t)\Rightarrow \boxed {x=2.40 cm\ sin(350 t)}$

At $x=1.20\ cm$

$1.20\ cm=2.40\ cm\ sin(350 t)$

$sin(350t)=0.5\Rightarrow sin(350t)=sin(\pi/6)$

$\Rightarrow 350t=\pi/6\Rightarrow t=\frac{\pi}{6\times 350}\ seconds$

$t=0.00150\ seconds = 1.50 \ ms$

At $x=2.40\ cm$ ,

$2.40\ cm=2.40\ cm\ sin(350t)\\ sin(350t)=1\\ sin(350t)=sin(\pi/2)\\ 350t=\pi/2\\ t=\frac{\pi}{350\times 2}\\ t=0.00449\ seconds = 4.49\ ms$

Answer: The position of the tip at any time is given by $x=2.40cm\ sin(350t)$

Time taken to travel from x=0 to x=1.20 cm is 1.50 ms.

Time taken to travel from x=0 to x=2.40 cm is 4.49 ms