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]
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" .
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