Question 302: Oscillations

Question

A simple pendulum of length '$$l$$' and a bob of mass '$$\mathrm{m}$$' is executing S.H.M. of small amplitude '$$A$$'. The maximum tension in the string will be ($$\mathrm{g}=$$ acceleration due to gravity)

Accepted Answer

Tension in the string is given as,
$$\begin{aligned}
& \mathrm{T}^{\prime}=\mathrm{mg} \cos 	heta+\frac{\mathrm{mv}^2}{l} \
& \mathrm{~T}_{\max }^{\prime}=\mathrm{mg}+\frac{\mathrm{mv}^2}{l} \
\end{aligned}$$
Now, time period $$\mathrm{T}=2 \pi \sqrt{\frac{l}{g}}$$
$$\omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{\frac{\mathrm{g}}{l}}$$
Maximum Velocity is given as $$\mathrm{v}_{\max }=A \omega$$
Substituting the values
$$\begin{aligned}
\mathrm{T}_{\max }^{\prime} & =\mathrm{mg}+\frac{\mathrm{m}(\mathrm{A} \omega)^2}{l} \
& =\mathrm{mg}+\frac{\frac{\mathrm{mA}^2 \mathrm{~g}}{l}}{l} \
& =\mathrm{mg}+\frac{\mathrm{mA}^2 \mathrm{~g}}{l^2} \
& =\mathrm{mg}\left(1+\left(\frac{\mathrm{A}}{l}ight)^2ight)
\end{aligned}$$

Answer

$$2 \mathrm{~mg}$$

Answer

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)^2\right]$$

Answer

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]^2$$

Answer

$$m g\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]$$

Oscillations

A simple pendulum of length '$l' and a bob of mass '\mathrm{m}' is executing S.H.M. of small amplitude 'A'. The maximum tension in the string will be (\mathrm{g}=$ acceleration due to gravity)

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