Question

Using equation x = Ae^-bt/2m t cos (ω't + φ) and assuming φ = 0 at t = 0, find the expression for acceleration at t = 0

• A. $A\left\lbrack \frac{k}{m} - \frac{b^{2}}{4m^{2}} \right\rbrack$
• B. $A\left\lbrack \frac{k}{m} + \frac{b^{2}}{4m^{2}} \right\rbrack$
• C. $A\left\lbrack \frac{b^{2}}{4m^{2}} - \frac{k}{m} \right\rbrack$
• D. $A\left\lbrack \frac{b}{2m} - \frac{k}{m} \right\rbrack$

Answer 1

Answer: (C)

$A\left\lbrack \frac{b^{2}}{4m^{2}} - \frac{k}{m} \right\rbrack$

Answer 2

$\frac{dx}{dt} = \frac{- b}{2m}Ae^{- bt/2m}\cos(\omega't) - Ae^{- bt/2m}\omega\sin(\omega't)$ $\frac{d^{2}x}{dt^{2}} = - \frac{- Ab}{2m}\left\lbrack \frac{- b}{2m}e^{- bt/em}\cos(\omega't) - e^{- bt/em}\omega'\sin(\omega't) \right\rbrack + \frac{bA}{2m}\omega'e^{bt/2m}\sin(\omega't + \varphi) - A\omega^{'2}e^{- bt/2m}\cos(\omega't + \varphi)\left. \ \frac{d^{2}x}{dt^{2}} \right|_{r = 0}$

= - A'ω² + A (b²/4m²) = $A\left\lbrack \frac{b^{2}}{4m^{2}} - \frac{k}{m} \right\rbrack$