## Question

A particle executes SHM of amplitude 25 cm and time periods 3 s. what is the minimum time required for the particle to move between two points located at 12.5 cm on either side of the mean position?

### Solution

0.5 s

Let A and B be the two extreme positions of the particle with *O* as the mean position. Displacements to the right of *O* are taken as positive while those to the left of *O* are taken as negative

Let the displacement of the particle is SHM be given by

..(i)

let us suppose that at time *t* = 0, the particle is at extreme position B. Setting *x* = *A* and *t* = 0 in Eq. (i) we have

*A* = *A* sin Ï•

Giving

Where A = 25 cm.

Now let us say that the particle reaches point *C* at *t* = *t*_{1} and point *D* at *t* =*t*_{2}. At *C*, the displacement *x*(*t*_{1}) = + 12.5 cm and at *D*, it is *x*(*t*_{2}) = –12.5 cm (see fig.) so from (ii) we have

Notice that cos ω*t*_{2} = –0.5 even for This value of *t*_{2} does not correspond to the minimum time because this is the time at which the particle, moving to left, reaches *A* and then returns to D.

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