Explanation
Angular frequency, \omega =\sqrt{\dfrac{k}{m}}\ \ \ \Rightarrow \ \dfrac{2\pi }{T}=\sqrt{\dfrac{k}{m}}
\Rightarrow \ k=m{{\left( \dfrac{2\pi }{T} \right)}^{2}} where, T is time period.
Net Spring constant when two spring is connected in parallel.
k={{k}_{1}}+{{k}_{2}}
\Rightarrow m{{\left( \dfrac{2\pi }{T} \right)}^{2}}=m{{\left( \dfrac{2\pi }{{{T}_{1}}} \right)}^{2}}+m{{\left( \dfrac{2\pi }{{{T}_{2}}} \right)}^{2}}
\Rightarrow \dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T}_{1}}^{2}}+\dfrac{1}{{{T}_{2}}^{2}}
Given that,
Amplitude = A
Time period = T
Average velocity in time period = \dfrac{T}{4}
The displacement equation of SHM
x=A\sin \omega t
We know that,
v=\dfrac{dx}{dt}
v=\dfrac{d\left( A\sin \omega t \right)}{dt}
v=A\omega \cos \omega t
Now, the average velocity is
<v{{>}_{0\to \dfrac{T}{4}}}=\dfrac{\int\limits_{0}^{\dfrac{T}{4}}{vdt}}{\int\limits_{0}^{\dfrac{T}{4}}{dt}}
<v>\,=\dfrac{\int\limits_{0}^{\frac{T}{4}}{A\omega \cos \omega t}}{\dfrac{T}{4}}
<v>=\dfrac{4A\omega }{T}\int\limits_{0}^{\frac{T}{4}}{\cos \omega tdt}
<v>=\dfrac{4A\omega }{T}\left[ \frac{\sin \omega t}{\omega } \right]_{0}^{\dfrac{T}{4}}
<v>=\dfrac{4A}{T}\left[ \sin \dfrac{2\pi }{T}\times \dfrac{T}{4}-0 \right]
<v>=\dfrac{4A}{T}
Hence, the average velocity is \dfrac{4A}{T}
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