Question

For a particle executing simple harmonic motion the amplitude is A and time period is T. The maximum speed will be​

Answer 1

Explanation:

​ The maximum possible average velocity = maximum displacement / time

 i.e. Vavg =   dmax /T/4

​ here T→2π

    T/4 = π /2

∴ θ =  π/4

∴ dmax = PQ = 2Asinθ

== 2asin π/4

∴ dmax == 2A/$\sqrt{2}$ = $\sqrt{2} A$

​ ∴ Vavg = $\frac{\sqrt{2}A }{T/4}$ = $\frac{4\sqrt{2} A}{T}$

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Answer 2

maximum speed=Aω  A=amplitude  and ω=angular frequency

time period T=2π/ω

                     ω=2π/T

maximum speed = 2πA/T