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PhysicsSHMCalculation of time periodMedium2 minQB

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A particle at the end of a spring executes simple harmonic motion with a period $t_{1}$ , while the corresponding period for another spring is $t_{2}$ . If the period of oscillation with the two springs in series is T , then

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Answer:

Solution:

$t_{1} = 2 π \frac{m}{k _{1}}$

$\Rightarrow k_{1} = \frac{4 π ^{2} m}{t _{1}^{2}}$ and $k_{2} = \frac{4 π ^{2} m}{t _{2}^{2}}$

$\frac{1}{k} = \frac{1}{k _{1}} + \frac{1}{k _{2}}$

$= \frac{t _{1}^{2} + t _{2}^{2}}{4 π ^{2} m}$

$T = 2 π m \frac{1}{k}$

$= 2 π m \frac{t _{1}^{2} + t _{2}^{2}}{4 π ^{2} m}$

$T^{2} = t_{1}^{2} + t_{2}^{2}$

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

⏱ 2mℹ️ Source: QB

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