Question

The magnitude of average acceleration in half time period in a simple harmonic motion is

(A) 2 ω2A /π
(B) ω2A /2 π
(C) ω2A /√2 π
(D) Zero

Answer 1

Option A - 2ω²A/π

Step-by-step explanation:

The displacement of simple harmonic motion is usually given as;

x = Asinωt

Where;

A is amplitude

ω is angular frequency

Now, average acceleration in half time period is given as;

a_av = d²x/dt² = -ω²A sinωt

Thus;

a_av = ∫a(t)dt/(∫dt) with both numerator and denominator between the boundaries of T/2 and 0.

Full expression gives;

(2/T)(ω²A∫sin(ωt) dt between the boundaries of T/2 and 0.

Now, integrating gives;

(2/T)(ω²A × 1/ω)[-cos(ωt)] between the boundaries of T/2 and 0.

Plugging in the boundary conditions gives us;

(2/T)(ω²A × 1/ω)[-cos(ωT/2) + cos0]

Simplifying gives;

(2ωA/T)[-cos(ωT/2) + 1]

In simple harmonic motion, we know that ω = 2π/T

​Thus we have;

(2ωA/T)[-cos(2π/2) + 1]

(2ωA/T)[-cos(π) + 1]

This gives;

(2ωA/T)[1 + 1]

= (4ωA/T)

Now, from earlier ω = 2π/T

Thus, T = 2π/ω

So, we have;

a_av = (4ωA/(2π/ω))

a_av = 2ω²A/π