Question

A particle oscillates in straight line simple harmonically with period 8 second and amplitude 4\[\sqrt 2\] m. Particle starts from mean position. The ratio of the distance travelled by it in 1^st second of its motion to that in 2^nd second is ______. (sin 45° = 1/\[\sqrt 2\], sin \[\frac {π}{2}\] = 1)

विकल्प

• 1:8

• 1:4

• 1:2

• 1:(\[\sqrt 2\] - 1)

Answer 1

A particle oscillates in straight line simple harmonically with period 8 second and amplitude 4\[\sqrt 2\] m. Particle starts from mean position. The ratio of the distance travelled by it in 1^st second of its motion to that in 2^nd second is 1:(\[\sqrt 2\] - 1).

Explanation:

For a particle executing SHM from mean position, x = A sin ωt

At t = 1 s,

x₁ = A sin\[\left({\frac{2\pi}{\mathrm{T}}}\mathbf{t}\right)\] = A sin \[\left(\frac{2\pi}{8}\times1\right)\]

= A sin 45° = \[\frac {A}{\sqrt 2}\]

t = 2 s,

x₂ = A sin\[\left({\frac{2\pi}{\mathrm{T}}}\mathbf{t}\right)\] = A sin \[\left(\frac{2\pi}{8}\times2\right)\]

= A sin 90° = A

\[\therefore\] Distance covered in 2^nd second,

x = x₂ - x₁ = A - \[\frac {A}{\sqrt 2}\] = A\[\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)\]

\[\therefore\] Required ratio,

\[\frac {x_1}{x}\] = \[\frac{\mathrm{A}/\sqrt{2}}{\mathrm{A}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}\] = \[\frac{4\sqrt{2}/\sqrt{2}}{4\sqrt{2}\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)}\] = \[\frac {1}{\sqrt 2 - 1}\]