Question 345: Oscillations

Question

A particle executes S.H.M. of period $$\frac{2 \pi}{\sqrt{3}}$$ second along a straight line $$4 \mathrm{~cm}$$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is

Accepted Answer

$$\begin{aligned} & \mathrm{T}=\frac{2 \pi}{\sqrt{3}} \mathrm{~s}, \mathrm{ZA}=4 \mathrm{~cm} \ & 	herefore \mathrm{A}=2 \mathrm{~cm}=2 	imes 10^{-2} \mathrm{~m} \ & 	ext { Acceleration, } a=\omega^2 \mathrm{x} \ & 	ext { Velocity, } v=\omega \sqrt{A^2-x^2} \ & 	herefore \quad \omega^2 \mathrm{x}=\omega \sqrt{A^2-\mathrm{x}^2} \ & \omega x=\sqrt{A^2-x^2} \ & \omega^2 x^2=A^2-x^2 \ & 	herefore \quad 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(	herefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}ight) \ & 	herefore \quad 4 \mathrm{x}^2=\mathrm{A}^2 \ & \mathrm{c}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm} \ & \end{aligned}$$

Answer

$$2 \mathrm{~cm}$$

Answer

$$1 \mathrm{~cm}$$

Answer

$$4 \mathrm{~cm}$$

Answer

$$3 \mathrm{~cm}$$

Oscillations

A particle executes S.H.M. of period $\frac{2 \pi}{\sqrt{3}} second along a straight line 4 \mathrm{~cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is

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