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A particle is executing SHM of amplitude r. AT a distance s from the mean position, the particle receives a blow in hte direction opposite to motion which instantaneously doubles the velocity. find the new amplitude.

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7 years ago

```							Simple Harmonic Motion: A body is said to be in simple harmonic motion such that, atany point its acceleration is directed always towards the mean position and it is directlyproportional to its displacement from the mean position.Consider a particle moving along a circle of radius A with uniform angular velocity ω . LetO be the centre of the circle and XX 1 and YY 1 are two mutually perpendicular diameters ofthe circle as shown in figure. Let the particle is at P at any instant of time t Let PN be thenormal drawn on to the diameter YY 1 from P. As P moves on the circumference of thecircle, N moves on the diameter YY 1 . Let the angular displacement of the particle P is∠XOP = θThe displacement of N with respect to the fixed point O in the path is given byON = Y = A sin θ (or) Y = A sin ω t (Q θ = ω t ) ................. (1)The acceleration of N ( aN ) is equal to the component of the centripetal acceleration of Pparallel to the diameter YY 1 .The centripetal acceleration of P along PO is given bya p = Aω 2 .............(2)The centripetal acceleration parallel to NO.aN = aP sin θ .............(3)From equations (2) and (3) , we get,aN = Aω 2 sin ωtUsing equation (1) ,aN = −ω 2 yThe negative sign indicates that aN and Y are in opposite directions. Since ‘ ω ’ is aconstant then we can write aN α − yi.e, the acceleration of N is directly proportional to the displacement in magnitude but inopposite direction, always towards the fixed point ‘O’ in the path.Hence the projection of uniform circular motion on t the diameter is simple harmonic.(b) A = 0.04 m ; v = 50Hz ;φ =π /3
y = A sin (ω t + φ ) = A sin ( 2π vt + φ )y = 0.04 sin (100π t + π / 3 ) m.
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7 years ago
```							 before the blow, we know the total energy of the system was 1/2 k r^2 since when the particle is at maximum amplitude, all the mechanical is in the form of elastic PE which is equal to 1/2 k x^2 where x is the displacement from equilibrium just before the blow, the PE of the particle was 1/2 k s^2, so its KE at the time was 1/2 k r^2 - 1/2 k s^2 = 1/2 k(r^2-s^2) if the blow doubled the velocity, the KE quadrupled (since KE depends on v^2), so the KE immediately after the blow was 2 k(r^2 - s^2) now, the PE is still 1/2 k s^2, so the total energy of the system was total energy = total PE + total KE = 1/2 k s^2 + 2 k(r^2 - s^2) = 2 k r^2 - 3/2 k s^2 since this is the total energy, it must be equal to the new total energy when the particles is at the farthest point, which we call X. The PE at X is 1/2 k X^2 and this equals total energy so 1/2 k X^2 = 2 k r^2 - 3/2 k s^2 divide through by 1/2 k and obtain X^2 = 4 r^2 - 3 s^2 and X = Sqrt[4 r^2 - 3s^2]
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3 years ago
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