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Physics - SHM Question with Solution | TestHub

PhysicsSHMMiscellaneousMedium2 minPYQ_2014

PhysicsMediumsingle choice

A mass of 1 kg is suspended from a spring of force constant 400 N , executing SHM total energy of the body is 2 J , then maximum acceleration of the spring will be

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Solution:

$T = 2 π \frac{m}{R} = 2 π \frac{1}{400} = \frac{2 π}{20}$ Frequency $n = \frac{20}{2 π}$ $ω = 2 πn = 2 π \times \frac{20}{2 π} = 20$ Total energy $E = \frac{1}{2} m ω^{2} r^{2} 2 = \frac{1}{2} \times 1 \times 400 \times r^{2}$ $r = \frac{1}{10} m Acceleration = ω^{2} r = 400 \times \frac{1}{10} = 40 m / s^{2}$

Stream:JIPMERSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2014

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