Simple harmonic motion

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Question

Figure shows the kinetic energy K of a simple pendulum versus its angle $θ$ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meter $(g = 10 m / s^{2})$ is _____________.

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Solution

$Given that from graph \frac{1}{2} {mV}_{m}^{2} = 15 \times 10^{- 3}$
$\begin{array}{l} V_{m} = \sqrt{0 .150} m / s \\ Aω = \sqrt{0 .150} m / s \\ {Lθ}_{m} \cdot \sqrt{\frac{g}{L}} = \sqrt{0 .150} m / s \\ \Rightarrow \sqrt{gL} = \frac{\sqrt{0 .150}}{100 \times 10^{- 3}} \\ \Rightarrow L = \frac{0 .150}{0 .1} = 1 .5 m \end{array}$

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Simple harmonic motion

Remember concepts with our Masterclasses.

Figure shows the kinetic energy K of a simple pendulum versus its angle θ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meterg=10m/s2 is _____________.

Given that from graph 12mVm2=15×10−3Vm=0.150m/sAω=0.150m/sLθm⋅gL=0.150m/s⇒ gL=0.150100×10−3⇒ L=0.1500.1=1.5m

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Figure shows the kinetic energy K of a simple pendulum versus its angle $θ$ from the vertical. The pendulum bob has mass 0.2 kg. The length of the pendulum in meter $(g = 10 m / s^{2})$ is _____________.