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PhysicsSHMCalculation of time periodEasy2 minPYQ_2020

PhysicsEasysingle choice

The kinetic energy of a particle executing S.H.M. is $16 J$ when it is at its mean position. If the amplitude of oscillations is $25 cm$ , and the mass of the particle is $5.12 kg$ , then the time period of the oscillation is

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Answer:

Solution:

When the particle is in its mean position, the kinetic energy will be maximum,

$K_{max} = \frac{1}{2} m ω^{2} a^{2}$ and $ω = \frac{2 π}{T}$

$\Rightarrow 16 = \frac{1}{2} \times 5.12 \times \frac{4 π^{2}}{T^{2}} \times {(0.25)}^{2}$

$\Rightarrow T^{2} = \frac{2.56 \times 4 π^{2} \times 0.25}{4} \times 0.25$

$\Rightarrow T = \frac{1.6 \times 2 π \times 0.25}{4} = \frac{π}{5} s$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2020

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