# 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.

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.

## 2 Answers

Simple Harmonic Motion: A body is said to be in simple harmonic motion such that, at

any point its acceleration is directed always towards the mean position and it is directly

proportional to its displacement from the mean position.

Consider a particle moving along a circle of radius A with uniform angular velocity ω . Let

O be the centre of the circle and XX 1 and YY 1 are two mutually perpendicular diameters of

the circle as shown in figure. Let the particle is at P at any instant of time t Let PN be the

normal drawn on to the diameter YY 1 from P. As P moves on the circumference of the

circle, N moves on the diameter YY 1 . Let the angular displacement of the particle P is

∠XOP = θ

The displacement of N with respect to the fixed point O in the path is given by

ON = 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 P

parallel to the diameter YY 1 .

The centripetal acceleration of P along PO is given by

a 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 ωt

Using equation (1) ,

aN = −ω 2 y

The negative sign indicates that aN and Y are in opposite directions. Since ‘ ω ’ is a

constant then we can write aN α − y

i.e, the acceleration of N is directly proportional to the displacement in magnitude but in

opposite 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.

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]