A particle is executing simple harmonic motio | vedclass

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(B) Given: Time period $T = 2 \ s$,Amplitude $A = 1 \ cm$,Total time $t = 12.5 \ s$.Number of cycles $n = \frac{t}{T} = \frac{12.5}{2} = 6.25$ cycles.

Vedclass · Accepted Answer

$25$

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(B) Given: Time period $T = 2 \ s$,Amplitude $A = 1 \ cm$,Total time $t = 12.5 \ s$.Number of cycles $n = \frac{t}{T} = \frac{12.5}{2} = 6.25$ cycles.In one complete cycle,the total distance covered is $4A = 4 	imes 1 \ cm = 4 \ cm$.For $6$ complete cycles,distance $D_1 = 6 	imes 4 \ cm = 24 \ cm$.For the remaining $0.25$ cycle (which is $\frac{T}{4}$),the particle moves from the mean position to the extreme position,covering a distance of $A = 1 \ cm$.Total distance $D = 24 \ cm + 1 \ cm = 25 \ cm$.After $6$ complete cycles,the particle returns to the mean position. In the remaining $0.25$ cycle,it moves from the mean position to the extreme position $(A = 1 \ cm)$.Thus,the displacement $d = 1 \ cm$.Therefore,$\frac{D}{d} = \frac{25 \ cm}{1 \ cm} = 25$.

$A$ particle is executing simple harmonic motion with time period $2 \ s$ and amplitude $1 \ cm$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5 \ s$,then $\frac{D}{d}$ is:

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