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A body executing SHM has a velocity of 7m/s at a distance of 4m from the mean position and 5m/s at a distance of 5m from the mean position . Find the amplitude and time period of SHM |
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Answer» Answer: Explanation: Equation of the velocity, v²=ω²(a²−y²) where, V = velocity, ω = angular frequency a = amplitude y = displacement from the mean position => Here, A body EXECUTING SHM has a velocity of 7m/s at a DISTANCE of 4m from the mean position 7² = ω²(a²- 4²) 49=ω²(a²−16) ...(1) => A body executing SHM has a velocity of 5m/s at a distance of 5m from the mean position. 5²=ω²(a²−5²) 25=ω²(a²−25) ...(2) => dividing equation (2) by equation (1), we get, 25/49 = a² - 25 / a² - 16 25 (a² - 16) = 49 (a² - 25) 25a² - 400 = 49a² - 1,225 1,225 - 400 = 49a² - 25a² 825 = 24a² a² = 34.375 a = 5.86 => now putting the VALUE of a in eqn (1), we get 49=ω²(34.375−16) 49 = ω² * 18.375 ω² = 49 / 18.375 ω² = 2.666 ω = 1.63 rad/s . => Time period of SHM: T = 2π / ω = 2 * 3.14 / 1.63 = 3.85 s Thus, the amplitude and time period of SHM is 1.63 rad/s and 3.85 s. |
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