Answer to Question #309400 in Mechanics | Relativity for Lipika

Question

The displacement of an object executing simple harmonic oscillations is given by:

x = 0.02 sin 2π ( t + 0.01 ) m

Determine (a) amplitude of the oscillatory motion, (b) time period of oscillation , (c) maximum velocity, (d) maximum acceleration, and (e) initial displacement of the object.

Answer 1

Solution.

$x=0.02sin2\pi(t+0.01)=0.02sin(2\pi t+0.02\pi)m;$

$x=Asin(\omega t+\phi_0);$

$a)A=0.02m;$

$b)\omega=\dfrac{2\pi}{T}\implies T=\dfrac{2\pi}{\omega};$ $\omega=2\pi s;$

$T=\dfrac{2\pi}{2\pi}=1s;$

$c)v_{max}=A\omega;$

$v_{max}=0.02\sdot2\sdot3.14=0.1256m/s;$

$d) a_{max}=A\omega^2;$

$a_{max}=0.02\sdot(2\sdot 3.14)^2=0.789 m/s^2;$

$e)x_0=0.02sin0.02\pi=0.0011m;$

Answer: $a)A=0.02m;$

$b)T=1s;$

$c)v_{max}=0.1256m/s;$

$d)a_{max}=0.789 m/s^2;$

$e)x_0=0.0011m.$