MHT-CET : Physics Entrance Exam

### MHT - CET : Physics - Simple Harmonic Motion Page 2

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We have,

d2x

=

-k

x

dt2

m

 i.e., a = -k x m

i.e., a = -w2x

 where, k = w2 m

The dimensions of both are [M°L°T
-2]

 A solution of differential equation d2x + w2x = 0 dt2

 d2x + w2x = 0 dt2

 \ d2x = -w2x dt2

 \ dn = - w2x dt

...(1)

 \ d2x = d ( dx ) dt2 dt dt

 = dn dt

 Now, dn = dn ´ dx = dn .n … (2) dt dx dt dx

 where n = dx is the velocity of the particle dt

 From (1) and (2); n dn = -w2x dx

\ ndn = -w2xdx

 On integrating, n2 = -w2x2 + C, where C is the constant of integration. 2 2

When
n = 0, x = ± A

 \ C = w2A2 2

 \ v2 = -w2x2 + w2A2 2 2 2

\ n = ± w

This is the expression for the velocity of the particle in terms of its displacement.

 Now, v = dx = w (considering only the positive root) dt

 \ dx = wdt

 On integration, sin-1 ( x ) = (wt + a) where a is the constant of integration A

which depends on the initial conditions.

\ x = A sin (wt + a)
This is the expression for the displacement of the particle.

Alternate Expressions For Velocity and Acceleration.
Since
x = A sin (wt + a)

 Velocity, n = dx = Aw cos (wt + a) dt

 Acceleration, a = dn = - Aw2 sin (wt + a) = -w2x dt

Maximum and Minimum Values of
x, n and a

 a) x = A sin (wt + a) \ |x max| = A when sin (wt + a) = ± 1 x min = 0 when sin (wt + a) = 0 b) n = A w cos (wt + a) \ |nmax| = Aw when cos (wt + a) = ± 1 nmin = 0 when cos (wt + a) = 0 Also, n = ± w \ n = nmax = wA when n = 0 nmin = 0 when x = ± A c) a = -Aw2 sin (wt + a) \ |amax| = Aw2 when sin (wt + a) = ± 1 and amin = 0 when sin (wt + a) = 0 also a = -w2x \ |amax| = w2A when x = ± A and amin = 0 when x = 0

Expression for Time Period and Frequency

 Time period T = 2p , but w2 = k , where k is the force constant and w m

m is the mass of the particle.
\ T = 2p

 Also, since |a| = w2x, w2 = |a| x

 \ w =

\ w =

 T = 2p = 2p w

 Frequency n = 1 = w = 1 T 2p 2p

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