A particle is executing SHM. The time taken f | vedclass

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(A) In $SHM$,one full oscillation corresponds to a path length of $4A$ (where $A$ is the amplitude). We divide the path into $8$ equal parts of length

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$\frac{5x}{4}$

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(A) In $SHM$,one full oscillation corresponds to a path length of $4A$ (where $A$ is the amplitude). We divide the path into $8$ equal parts of length $A/2$ each. The time taken to travel these segments is shown in the diagram.For a displacement of $\frac{3}{8}$ of an oscillation from an extreme position,the particle travels from $x = A$ to $x = 0$ (which is $1/4$ of an oscillation) and then continues for another $1/8$ of an oscillation.The time taken is $T/4 + T/12 = T/3$.Given that this time is $x$,we have $T/3 = x$,which implies $T = 3x$.Now,for $\frac{5}{8}$ of an oscillation from the mean position $(x = 0)$,the particle travels $1/8 + 1/8 + 1/8 + 1/8 + 1/8 = 5/8$ of the path.The time taken is $T/12 + T/12 + T/12 + T/12 + T/12 = 5T/12$.Substituting $T = 3x$,the time is $5(3x)/12 = 15x/12 = 5x/4$.

$A$ particle is executing $SHM$. The time taken for $\left(\frac{3}{8}\right)^{\text{th}}$ of an oscillation from extreme positions is $x$. Then,the time taken for the particle to complete $\left(\frac{5}{8}\right)^{\text{th}}$ of an oscillation from the mean position is

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