Question

Acceleration of 3rd maxima w.r.t 3rd maxima on other side of central maxima at $t=3s$ is

• A. $0.02m{s}^{-2}\hat{i}$
• B. $0.03m{s}^{-2}\hat{i}$
• C. $10m{s}^{-2}\hat{i}$
• D. $0.6m{s}^{-2}\hat{i}$

Answer 1

Answer: (B)

$0.03m{s}^{-2}\hat{i}$

Position of 3rd maxima, $y=\frac{3\lambda D}{d}$

Differentiating twice, $a_3=\frac{2\lambda }{d}\frac{d^2}{d{t}^2}\left(D\right)$

As $D=1+\frac{1}{2}g{t}^2$

Thus $a_3=\frac{3\lambda g}{d}=.015m{s}^{-2}\hat{i}$ (for right side of central maxima)

Similarly $a_3^{\prime }=\frac{3\lambda g}{d}=-.015m{s}^{-2}\hat{i}$ (for left side of central maxima)

Now acceleration of 3rd maxima w.r.t 3rd maxima on other side ofcentral maxima, $a=.015m{s}^{-2}\hat{i}-(-.015)m{s}^{-2}\hat{i}⟹a=.03m{s}^{-2}\hat{i}$