Question

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Answer 1

❏ Question:-

@ A mass m is attached to a spring , is free to oscillate with angular frequency w in a horizontal plane without friction . The mass is pulled to a position distant $\sf{X_o}$ from the mean position and pushed to the mean position with a velocity $\sf{V_o}$ towards mean position. If theta be the Initial phase of its displacement , then the displacement equation is ,

(i) $\sf{\ \ {\cos\theta=\dfrac{V_o}{WX_o} }}$

(ii) $\sf{\ \ {\sin\theta=\dfrac{V_o}{WX_o} }}$

(iii) $\sf{\ \ {\cos\theta=\dfrac{1}{\sqrt{1+\dfrac{(V_o)^2}{(WX_o)^2}}} }}$

(iv) $\sf{\ \ {\sin\theta=\dfrac{1}{\sqrt{1+\dfrac{(V_o)^2}{(WX_o)^2}}} }}$

❏ Solution:-

For a Oscillating mass m the equation of displacement is given by ,

$\boxed{\large{\sf{\ \ {X= A\cos(Wt+\theta)}}}}$

Now at , time (t) = 0 , X=$\sf{X_o}$

$\implies\sf{\ \ {X_o= A\cos(W(0)+\theta)}}$

$\implies\boxed{\sf{\ \ {X_o= A\cos\theta}}}$ --------------(1)

Now, differentiating With respect to t ,

$\implies\sf{\ \ {\dfrac{dX}{dt}=\dfrac{d}{dt} \{A\cos(Wt+\theta)\}}}$

$\implies\sf{\ \ {V=-AW\sin(Wt+\theta)}}$

Now at , time (t) = 0 , V=$\sf{V_o}$

$\implies\sf{\ \ {V_o=-AW\sin(W.0+\theta)}}$

$\implies\sf{\ \ {V_o=-AW\sin\theta}}$

$\implies\boxed{\sf{\ \ {A\sin\theta=\dfrac{V_o}{W}}}}$ --------------(2)

Now from equation (1) & (2) ,

doing [(2)²+(1)²] , we get ,

$\implies\sf{\ \ {(A\sin\theta)^2+(A\cos\theta)^2=(\dfrac{V_o}{W})^2+{X^2}_{o}}}$

$\implies\sf{\ \ {A^2\sin^2\theta+A^2\cos^2\theta=(\dfrac{V_o}{W})^2+{X^2}_{o}}}$

$\implies\sf{\ \ {A^2(\sin^2\theta+\cos^2\theta)=\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}}$

$\implies\sf{\ \ {A^2(1)=\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}}$

$\implies\sf{\ \ {A^2=\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}}$

$\implies\boxed{\sf{\ \ {A=\sqrt{\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}}}}$

Hence from equation (1)

$\implies\sf{\ \ {X_o= A\cos\theta}}$

$\implies\sf{\ \ {\cos\theta=\dfrac{X_o}{A}}}$

$\implies\sf{\ \ {\cos\theta=\dfrac{X_o}{\sqrt{\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}}}}$

$\implies\sf{\ \ {\cos\theta=\dfrac{1}{\sqrt{\dfrac{\dfrac{{V^2}_{o}}{W^2}+{X^2}_{o}}{{X^2}_{o}}}}}}$

$\implies\sf{\ \ {\cos\theta=\dfrac{1}{\sqrt{\dfrac{{V^2}_{o}}{W^2{X^2}_{o}}+1}}}}$

$\implies\large{\boxed{\sf{\ \ {\cos\theta=\dfrac{1}{\sqrt{1+\dfrac{V^2_o}{W^2X_o^2}}}}}}}$

Option ➩ (iii) $\sf{\ \ {\cos\theta=\dfrac{1}{\sqrt{1+\dfrac{(V_o)^2}{(WX_o)^2}}} }}$