A simple harmonic motion is represented by α d t 2 d 2 x + β x = 0 . Its period is

The standard equation of SHM is: d t 2 d 2 x = − ω 2 x On comparing with equation: α d t 2 d 2 x + β x = 0 The angular frequency is given by: ω = α β . Therefore, the time period of SHM is: T = ω 2 π = 2 π β α

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PhysicsSHMCalculation of time periodHard2 minPYQ_2022

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A simple harmonic motion is represented by $α \frac{d ^{2} x}{d t ^{2}} + β x = 0$ . Its period is

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The standard equation of SHM is: $\frac{d ^{2} x}{d t ^{2}} = - ω^{2} x$ On comparing with equation: $α \frac{d ^{2} x}{d t ^{2}} + β x = 0$ The angular frequency is given by: $ω = \frac{β}{α}$ . Therefore, the time period of SHM is: $T = \frac{2 π}{ω} = 2 π \frac{α}{β}$

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

⏱ 2mℹ️ Source: PYQ_2022

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