Question

The initial phase of body executing SHM is π 4 then what will be its phase at the end of 10 oscillations ?​

Answer 1

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Answer 2

Given:

The initial phase of body executing SHM is π/4.

To find:

Phase after 10 oscillations?

Calculation:

Since initial phase is π/4, we can say:

• Equation of SHM will be :

$x = a \sin \bigg( \omega t + \dfrac{\pi}{4} \bigg)$

Now, initially, when t = 0 , phase is π/4.

Now, when 10 oscillations are complete, we can say t = 10(T) , where T is time period of SHM.

$\implies x = a \sin \bigg( 10\omega T+ \dfrac{\pi}{4} \bigg)$

$\implies x = a \sin \bigg( \dfrac{2\pi}{T} \times 10 T+ \dfrac{\pi}{4} \bigg)$

$\implies x = a \sin \bigg( 20\pi+ \dfrac{\pi}{4} \bigg)$

$\implies x = a \sin \bigg( \dfrac{\pi}{4} \bigg)$

Phase after 10 oscillations still remain π/4.