Question

A particle executes simple harmonic oscillation with
an amplitude a. The period of oscillation is T. The
minimum time taken by the particle to travel half of
the amplitude from the equilibrium position is​

Answer 1

Answer:

t = $\frac{T}{12}$

Explanation:

Heya friend,

We know,

y = a sinωt

where,

y = distance traveled from mean position

a = amplitude / maximum distance traveled  

t = time taken by the body at any particular instant

Now,

as per the question,

y = $\frac{a}{2}$

So,

$\frac{a}{2}$ = a sinωt

or,

sinωt = $\frac{1}{2}$

Now,  

sin ωt = sin π/6

So,

ωt = π/6

t = π / 6ω

Now, We know, ω = 2π / T

So,

t = π / 6(2π / T)

t = $\frac{T}{12}$

Thanks !

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