Question

Using equation $x=A{e}^{(\frac{-bt}{2m})}\cos ({\omega }^{\prime }t+\varphi )$ and assuming $\varphi =0$ at $t=0$, find the expression for acceleration at $t=0$.

• A. $A[\frac{k}{m}-\frac{b^2}{4m^2}]$
• B. $A[\frac{k}{m}+\frac{b^2}{4m^2}]$
• C. $A[\frac{b^2}{4m^2}-\frac{k}{m}]$
• D. $A[\frac{b}{2m}-\frac{k}{m}]$

Answer 1

The correct option is C $A[\frac{b^2}{4m^2}-\frac{k}{m}]$

$x=A{e}^{\frac{-bt}{2m}}\cos ({\omega }^{\prime }t+\varphi )\Rightarrow v=\frac{dx}{dt}=\frac{-b}{2m}A{e}^{\frac{-bt}{2m}}\cos ({\omega }^{\prime }t+\varphi )-A{e}^{\frac{-bt}{2m}}{\omega }^{\prime }\sin ({\omega }^{\prime }t+\varphi )\Rightarrow a=\frac{d^{2}x}{d{t}^2}=\frac{-Ab}{2m}[\frac{-b}{2m}{e}^{\frac{-bt}{2m}}\cos ({\omega }^{\prime }t+\varphi )-e^{\frac{-bt}{2m}}{\omega }^{\prime }\sin ({\omega }^{\prime }t+\varphi )]+\frac{bA}{2m}{\omega }^{\prime }{e}^{\frac{-bt}{2m}}\sin ({\omega }^{\prime }t+\varphi )$

$-A{\omega }^{\prime 2}{e}^{\frac{-bt}{2m}}\cos ({\omega }^{\prime }t+\varphi )\Rightarrow a(\varphi =0,t=0)=-A{\omega }^{\prime 2}+A\frac{b^2}{4m^2}=A[\frac{b^2}{4m^2}-\frac{k}{m}]$