Question

the maximum speed and acceleration of a particle executing SHM be 20 cm and 20 CM square respectively then the time period is​

Answer 1

Answer:

2π

Explanation:

Vmax = Aw

Amax= -A2(square) w

VmAx/Amax gives 1/w= 1

I.e Aw/Aw2=1/w therefore w =1

Time period= 2π/w hence time period is 2π/1=2π ...

Answer 2

The time period of the particle is $2\pi\ seconds$.

Explanation:

It is given that,

Maximum speed of the particle, $v_{max}=20\ cm$

Maximum acceleration of the particle, $a_{max}=20\ cm$

We know that, the maximum speed and acceleration of the particle in SHM is given by :

$v_{max}=A\omega$.........(1)

$a_{max}=A\omega^2$..........(2)

Dividing equation (1) and (2), we get :

$\dfrac{v_{max}}{a_{max}}=\dfrac{A\omega}{A\omega^2}$

$\dfrac{v_{max}}{a_{max}}=\dfrac{1}{\omega}$

Since, $\omega=\dfrac{2\pi}{T}$, T is the time period

$\dfrac{v_{max}}{a_{max}}=\dfrac{T}{2\pi}$

$T=\dfrac{2\pi \times v_{max}}{a_{max}}$

$T=\dfrac{2\pi \times 20}{20}$

$T=2\pi\ seconds$

So, the time period of the particle is $2\pi\ seconds$. Hence, this is the required solution.

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