Explanation
Using the expressions for velocity for SHM,
$$v_1 = \omega \sqrt{A^2 -x_1^2}$$$$v_2 = \omega \sqrt{A^2 -x_2^2}$$Squaring both,$$v_1 ^2 = \omega ^2 ( A^2 -x_1 ^2)$$ ....(3)$$v_2 ^2 = \omega ^2 ( A^2 -x_2 ^2)$$ ....(4)Subtracting (4) from (3), $$v_1 ^2 - v_2 ^2 = \omega ^2 ( x_2 ^2 - x_1 ^2) $$$$ \Rightarrow 4 \pi ^2 ( x_2 ^2 - x_1 ^2) = T^2 (v_1 ^2 - v_2 ^2 )$$$$ \Rightarrow T = 2\pi \sqrt{\dfrac{ x_2 ^2 - x_1 ^2}{v_1 ^2 - v_2 ^2}}$$ where we have used $$\omega = \dfrac{2\pi}{T}$$
HINT: DISTANCE IS DEFINED AS THE AREA COVERED BY AN OBJECT FROM MOVING FROM ONE POINT TO ANOTHER
STEP1:In fig (1) particles at a mean position.In fig(2) particles staring some displacement but crossing each other in opposite direction.
STEP2:Now if the displacement of two particle
$$\mathrm{x}_{1}=\mathrm{A} \sin (\omega \mathrm{t})$$
$$\mathrm{x}_{2}=\mathrm{A} \sin (\omega \mathrm{t})$$
For $$\mathrm{x}_{1}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} \ldots, \dfrac{5 \pi}{6} \ldots, \ldots$$
$$\mathrm{x}_{2}=\dfrac{\mathrm{A}}{2}, \omega \mathrm{t}=\dfrac{\pi}{6} ., \dfrac{5 \pi}{6}$$
From fig (2) we can say point $$A$$ and B corresponding $$\omega \mathrm{t}=\dfrac{5 \pi}{6}, \omega \mathrm{t}=\dfrac{\pi}{6}$$ So phase difference $$\left(\dfrac{5 \pi}{6}-\dfrac{\pi}{6}\right)=\dfrac{2 \pi}{3}$$
[N.B $$\rightarrow$$ Before starting motion of ' 2 '...particle ' 1 ' goes to right extreme end and while '1 returning to mean position particle '2' starts its motion and the point of equal displacements are A and B]
Number of oscillations made by a vibrating body of an medium in 1 sec is known as frequency of a wave. And one oscillation completed by a vibrating body in one second is said to be 1 Hz.
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