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(B) The equation of motion for a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.Given amplitude $A = 4\, cm$ and time period $T

Vedclass · Accepted Answer

$0.4$

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(B) The equation of motion for a particle in simple harmonic motion is given by $x = A \sin(\omega t)$.Given amplitude $A = 4\, cm$ and time period $T = 1.2\, s$.To find the time taken to move from $x = 2\, cm$ to $x = 4\, cm$:At $x = 2\, cm$,$2 = 4 \sin(\omega t_1) \implies \sin(\omega t_1) = 1/2 \implies \omega t_1 = \pi/6$.At $x = 4\, cm$,$4 = 4 \sin(\omega t_2) \implies \sin(\omega t_2) = 1 \implies \omega t_2 = \pi/2$.The time taken to go from $x = 2\, cm$ to $x = 4\, cm$ is $\Delta t = t_2 - t_1 = (\pi/2 - \pi/6) / \omega = (\pi/3) / (2\pi/T) = T/6$.Since the particle returns from $x = 4\, cm$ to $x = 2\, cm$,the time taken for the return journey is also $T/6$.Total minimum time $= T/6 + T/6 = T/3$.Given $T = 1.2\, s$,total time $= 1.2 / 3 = 0.4\, s$.

$A$ particle is performing simple harmonic motion along the $x$-axis with an amplitude of $4\, cm$ and a time period of $1.2\, s$. The minimum time taken by the particle to move from $x = 2\, cm$ to $x = +4\, cm$ and back again is given by .... $s$.

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