A) \[{{a}^{2}}{{T}^{2}}+4{{\pi }^{2}}{{v}^{2}}\]
B) \[aT/x\]
C) \[aT+2\pi n\]
D) \[aT/v.\]
Correct Answer: A
Solution :
\[x=A\,sin\,\omega t\] \[v=A\omega \,\cos \,\omega t\] \[a=-A\omega \text{ }\sin \,\omega \theta \] Hence, \[{{a}^{2}}{{T}^{2}}+4{{\pi }^{2}}{{v}^{2}}\] \[={{\sin }^{2}}\omega \theta (-A{{\omega }^{2}}){{\left( \frac{2\pi }{\omega } \right)}^{2}}4{{\pi }^{2}}{{(A\omega )}^{2}}\cos \theta \] \[4{{\pi }^{2}}{{(A\omega )}^{2}}[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta ]\] which is time independent. Option (2) however may not be regarded as the correct answer because it is not mentioned that\[x,\]the displacement is measured from equilibrium position.
You need to login to perform this action.
You will be redirected in
3 sec
