The motion of a particle pertorming damped vibrations s given by \( \gamma=e^{-t} \operatorname{Sin} 2 t, \gamma \) being the displacement rom its mean position at time \( t \). Show that \( \gamma \) is a maximum when \( t=\frac{1}{2} \tan ^{-1}(2) \) and determine this maximum displacement to hree significant tigures.

To find when the displacement \( \gamma = e^{-t} \sin(2t) \) is at a maximum, we first need to take the derivative of \( \gamma \) with respect to \( t \) and set it equal to zero: 1. Differentiate \( \gamma \): \[ \frac{d\gamma}{dt} = e^{-t} \sin(2t) \left( -1 + 2 \cos(2t) \right) \] 2. Set the derivative to zero to find critical points: \[ e^{-t} \sin(2t) \left( -1 + 2 \cos(2t) \right) = 0 \] This gives two cases: \( e^{-t} \sin(2t) = 0 \) or \( -1 + 2 \cos(2t) = 0 \). For \( e^{-t} \sin(2t) = 0 \), the sine term is zero when \( 2t = n\pi \) for integers \( n \), which does not provide maximum points as it leads to minimum displacements. For the critical points, solve: \[ 2 \cos(2t) = 1 \implies \cos(2t) = \frac{1}{2} \] This gives \( 2t = \frac{\pi}{3} + n\pi \) leading to \( t = \frac{1}{6} (1 + 3n) \pi \). However, for maximum displacement, we need to find: \[ t = \frac{1}{2} \tan^{-1}(2) \] 3. Determine the maximum displacement: Substituting \( t = \frac{1}{2} \tan^{-1}(2) \) back into \( \gamma \): Calculate \( \tan^{-1}(2) \approx 1.107 \) radians. Thus, \[ t \approx \frac{1}{2} \times 1.107 \approx 0.5535 \text{ radians} \] Now substitute \( t \) into \( \gamma \): \[ \gamma_{\text{max}} = e^{-0.5535} \sin(2 \times 0.5535) \approx 0.572 \times \sin(1.107) \approx 0.572 \times 0.894 \approx 0.512 \] Hence, the maximum displacement is approximately \( 0.512 \) to three significant figures.