Answer to Question #196359 in Mechanics | Relativity for TAWANDA SIMON

Question

At time t = 0, the displacement x(0) of the block in a linear oscillator like that of Fig. 3

is -8.50 cm. The block’s velocity v(0) then is -0.93 m/s, and its acceleration a(0) is 45.0

m/s2 . Determine,

(i) the angular frequency of this system,

(ii) phase constant and

(iii) amplitude of this system. [3,5,3]

Answer 1

Solution.

$t=0;$

$x(0)=-8.50cm=-0.085m;$

$v(0)=-0.93m/s;$

$a(0)=45.0m/s^2;$

$i) x(t)=Acos(\omega t+\phi_0);$

$v(t)=-A\omega sin(\omega t+\phi_0);$

$a(t)=-A\omega ^2cos(\omega t+\phi_0);$

$-0.085=Acos\phi_0\implies cos\phi_0=-\dfrac{0.085}{A};$

$-0.93=-A\omega sin\phi_0;$

$45=-A\omega^2cos\phi_0=\omega^2\sdot 0.085\implies \omega^2=529.4; \omega=23rad/s;$

$ii) -0.93=-23 Asin\phi_0;$

$-0.085=Acos\phi_0$;

$10.94=-23tan\phi_0\implies tan\phi_0=-0.475;\phi_0=2.7rad;$

$iii) A=\dfrac{-0.085}{cos\phi_0}=0.094m;$

Answer:$i) \omega=23rad/s;$

$ii)\phi_0=2.7rad;$

$iii)A=0.094m$ .