Question

The speeds of a particle performing linear SHM are 8 units and 6 units at respective displacements of 6 cm and 8 cm. Find its period and amplitude.

Answer 1

v₁ = 8 units,  v₂ = 6 units,

x₁ = 6cm, x₂ = 8cm

We know that,

v = ω`sqrt((A^2  x^2))`

∴ `v_1 /v_2 = 8/6 = sqrt(A^2 - 6^2)/sqrt(A^2 - 8^2)`

i.e.  `4/3 = sqrt(A^2 - 6^2)/(A^2 - 8^2)`

 ∴ `16/9 = (A^2 - 36)/(A^2 - 64)`

∴ 16 (A² - 64) = 9 (A² - 36)

∴ 16A² - 9A² = 1024 - 324

7A²  = 700

A² =  100

∴ A = 10 cm

Now, consider

∴ `v_1 = ω sqrt(A^2 - x_1^2`

`8 = (2pi)/T sqrt (10^2 - 6^2)    ...(∵ ω = (2pi)/T)`

`T = (2 xx 3.142)/8 xx8`

∴ T  = 6.284 sec.