সরল দোলন গতি

A particle execute S.H.M from the mean position. its amplitude is AA, its time period is "T'. At what displacement,its speed is half of its maximum speed.

হানি নাটস

For a SHM velocity, v=ωA2x2v=\omega \sqrt{A^2-x^2}. . . . . . .(1)

The maximum velocity is at x=0x=0

vm=ωAv_m=\omega A

Now, v=ωA2v=\dfrac{\omega A}{2}

From equation (1),

ωA2=ωA2x2\dfrac{\omega A}{2}=\omega \sqrt{A^2-x^2}

A24=A2x2\dfrac{A^2}{4}=A^2-x^2

x2=3A24x^2=\dfrac{3A^2}{4}

x=3A2x=\dfrac{\sqrt{3}A}{2}

The correct option is A.

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