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A particle is moving in SHM in a straight line. When the distance of the particle from equilibrium position has values `x_(1)` and `x_(2)` , the corresponding values of velocities are `u_(1)` and `u_(2)`. Show that time period of vibration is `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)` |
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Answer» As, `V^(2)=omega^(2)(a^(2)-y^(2)),` so `u_(1)^(2)=omega^(2)a^(2)-omega^(2)x_(1)^(2)` …(i) And `u_(2)^(2)=omega^(2)a^(2)-omega^(2)x_(2)^(2)` …(ii) Subtracting (ii) from (i), we get, `u_(1)^(2)-u_(2)^(2)=omega^(2)(x_(2)^(2)-x_(1)^(2))` or `u_(1)^(2)-u_(2)^(2)=(4pi^(2))/(T^(2))(x_(2)^(2)-x_(1)^2)` or `T=2pi[(x_(2)^(2)-x_(1)^(2))/(u_(1)^(2)-u_(2)^(2))]^(1//2)` |
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