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Q.

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

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a

9 m/s2

b

2 m/s2

c

4.5 m/s2

d

12 m/s2

answer is B.

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Detailed Solution

For a simple pendulum, maximum acceleration is given by, a=ω2A=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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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