Answer to Question #96967 in Classical Mechanics for Surej

Question

body of mass 10g is in SHM with a frequency of 40/π Hz and amplitude of 0.2m.
Find :
a) The maximum force acing on the body. b) the maximum velocity and the
maximum acceleration.
c) the acceleration and the velocity at the point 0.1m from the equilibrium position.

Answer 1

Dependence of the coordinate (displacement relative to the equilibrium position) is given by the expression

$x(t) = A\sin (\omega t + \varphi )$

Where $\omega$ is a angular frequency, $\varphi$ - the initial phase and $A$ - amplitude. The angular frequency in our case are

$\omega = 2\pi \nu = 2\pi \cdot {{40} \over \pi }[{\rm{Hz}}] = 80[{{\rm{1}} \over {\rm{s}}}]$

Part a):

The magnitude of the force can be found by Newton's second law $\left| {\vec F} \right| = m\left| {{{{d^2}\vec x} \over {d{t^2}}}} \right|$ . Let's differentiate $x(t)$ with respect to $t$ twice

$\dot x(t) = A\omega \cos (\omega t + \varphi )$ $\ddot x(t) = - A{\omega ^2}\sin (\omega t + \varphi )$

Then, we need to maximize the expression $\left| {\vec F} \right| = Am{\omega ^2}\left| {\sin (\omega t + \varphi )} \right|$

It's obvious, that the $\max (\sin (\omega t + \varphi )) = 1$ , then

${\left| {\vec F} \right|_{\max }} = Am{\omega ^2}$

Let's calculate

${\left| {\vec F} \right|_{\max }} = 0.2[{\rm{m}}] \cdot 10 \cdot {10^{ - 3}}[{\rm{kg}}] \cdot {80^2}[{{\rm{1}} \over {{{\rm{s}}^{\rm{2}}}}}] = 12.8[{\rm{N}}]$

Part b):

The maximum acceleration is (Newton's second law)

${\left| {\vec a} \right|_{\max }} = {{{{\left| {\vec F} \right|}_{\max }}} \over m} = A{\omega ^2}$ ${\left| {\vec a} \right|_{\max }} = 0.2[{\rm{m}}] \cdot {(80)^2}[{{\rm{1}} \over {{{\rm{s}}^{\rm{2}}}}}] = 1280[{{\rm{m}} \over {{{\rm{s}}^{\rm{2}}}}}]$

The maximum of velocity can be found maximizing the first derivative

$\left| {\vec v} \right| = A\omega \left| {\cos (\omega t + \varphi )} \right|$

Again, $\max (\left| {\cos (\omega t + \varphi )} \right|) = 1$ and

${\left| {\vec v} \right|_{\max }} = A\omega$ ${\left| {\vec v} \right|_{\max }} = 0.2[{\rm{m}}] \cdot 80[{{\rm{1}} \over {\rm{s}}}] = 16[{{\rm{m}} \over {\rm{s}}}]$

Part c):

We need to find the such time $t$ that $x(t)=0.1[{\rm{m}}]$. Let $\varphi =0$ and then

Then

${{x(t)} \over A} = \sin (\omega t)$ $t = {1 \over \omega }\arcsin {{x(t)} \over A}$

In our case

$t = {1 \over {80[{{\rm{1}} \over {\rm{s}}}]}}\arcsin ({{0.1[{\rm{m}}]} \over {0.2[{\rm{m}}]}}) = {1 \over {80[{{\rm{1}} \over {\rm{s}}}]}}{\pi \over 6} = {\pi \over {480}}[{\rm{s}}]$

Then the velocity at this time can be calculates using the expression for the first derivative

$\left| {v({\pi \over {480}})} \right| = 0.2[{\rm{m}}] \cdot 80[{{\rm{1}} \over {\rm{s}}}]\left| {\cos (80[{{\rm{1}} \over {\rm{s}}}] \cdot {\pi \over {480}}{\rm{[s]}})} \right| = 16[{{\rm{m}} \over {\rm{s}}}] \cdot {{\sqrt 3 } \over 2} = 8\sqrt 3 [{{\rm{m}} \over {\rm{s}}}]$

For the acceleration we use the second derivative

$\left| {a({\pi \over {480}})} \right| = \left| { - 0.2[{\rm{m}}] \cdot {{80}^2}[{{\rm{1}} \over {{{\rm{s}}^2}}}]} \right|\left| {\sin (80[{{\rm{1}} \over {\rm{s}}}] \cdot {\pi \over {480}}{\rm{[s]}})} \right| = 1280[{{\rm{m}} \over {{{\rm{s}}^2}}}] \cdot {1 \over 2} = 640[{{\rm{m}} \over {\rm{s}}}]$