Question

The maximum speed and acceleration of a particle executing simple harmonic motion are 10 cm/s and 50 cm/s². Find the position(s) of the particle when the speed is 8 cm/s.

Answer 1

Maximum speed in the S.H.M. is given by the formula,

Vmax = ±ωA

Maximum acceleration is given by the relation,

Amax. = -ω²A

Dividing both the equations,

Vmax./Amax. = 1/ω

ω = 5

Now, V max. = ωA

10 = 5A

∴ A = 2 cm.

____________________________

The position of the particle is now x = Asin⍵t  [Taking x=0 at t=0]

∴ x = 2sin5t

Also,  v = A⍵cos⍵t

v = 2 × 5 cos5t,

Now, v = 8 cm/s,

8 = 10cos5t

cos5t = 8/10 =0.8

∴sin5t = √(1-cos²5t)

sin5t = √(1-0.64)

sin5t   =√0.36

sin5t   = ±0.6

∴ For v = 8 cm/s,  we have

x = 2 × (± 0.6 ) cm      

x = ± 1.2 cm.

Hope it helps.

Answer 2

$<i>$vmax = 10 cm/sec.

⇒ rω = 10

⇒ ω^2 = 100/r^2 ….(1)

Amax = ω^2r = 50 cm/sec

⇒ ω^2 = 50/y = 50/r …..(2)

∴ 100/r^2 = 50/r

⇒ r = 2 cm.

∴ ω = √(100/r^2 ) = 5 sec^2

Again, to find out the positions where the speed is 8m/sec,

v^2 = ω^2 (r2 – y2)

⇒ 64 = 25 ( 4 – y^2)

⇒ 4 – y^2 = 64/25

⇒ y^2 = 1.44

⇒ y = √1.44

⇒ $<b>$y = ± 1.2 cm from mean position.