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A particle of mass m= 5g is executing simple harmonic motion with an amplitude `0.3`m and time period `pi//5` second. The maximum value of force acting on the particle isA. 5 NB. 4 NC. `0.5` ND. `0.15`N |
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Answer» Correct Answer - D We know maximum acceleratiom, `a_("max")=omega^(2)A=(4pi^(2))/(T^(2))A` `=(4pi^(2))/(((pi)/(5))^(2))xx0.3=30m//s^(2)` Maximum force, `F_("max")=ma_("max")=(.5)/(1000)xx30=0.15N` |
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