Maximum acceleration of the bob of a simple pendulum oscillating with amplitude A is $$6m/s2$$ and that of another pendulum having same amplitude ‘A’ is $$3$$ $$m/s2$$. A third pendulum is formed by joining the strings of the pendulum and its bob is allowed to oscillate with same amplitude ‘A’. Then maximum acceleration of this bob is

Question

Maximum acceleration of the bob of a simple pendulum oscillating with amplitude A is $$6m/s2$$ and that of another pendulum having same amplitude ‘A’ is $$3$$ $$m/s2$$. A third pendulum is formed by joining the strings of the pendulum  and its bob is allowed to oscillate with same amplitude ‘A’. Then maximum acceleration of this bob is

Infinity Learn · Accepted Answer

For a simple pendulum, maximum acceleration is given by, $$a=$$ω$$2A=gl2A$$⇒$$l=gAa$$∴ Length of the third pendulum $$l3=l1+l2=gAa1+gAa2$$∴ Maximum acceleration of the bob of a third pendulum is, $$a3=gAl3=gAgAa1+gAa2=a1a2a1+a2$$⇒$$a3=6$$×$$36+3m/s2=2$$ $$m/s2$$

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