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Standard VI

Physics

Speed

Figure shows ...

Question

Figure shows an oscillating pendulum.

Time taken by the bob to move from A to C is $t_{1}$ and from C to O is $t_{2}$ . The time period of this simple pendulum is

(t_{1} + t_{2})

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2 (t_{1} + t_{2})

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3 (t_{1} + t_{2})

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4 (t_{1} + t_{2})

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Solution

The correct option is D $4 (t_{1} + t_{2})$
The total time taken by the bob to move from A to O is $(t_{1} + t_{2}),$
which is $\frac{1}{4^{t h}}$ time of one full cycle of the pendulum.
Therefore, time period of pendulum i.e. time taken by pendulum to complete one oscillation from A to B and back to A will be $4 (t_{1} + t_{2})$ .

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Figure shows an oscillating pendulum. Time taken by the bob to move from A to C is t1 and from C to O is t2. The time period of this simple pendulum is

The correct option is D 4 (t1 + t2)The total time taken by the bob to move from A to O is (t1+t2), which is 14th time of one full cycle of the pendulum. Therefore, time period of pendulum i.e. time taken by pendulum to complete one oscillation from A to B and back to A will be 4 (t1 + t2).

Figure shows an oscillating pendulum.

Time taken by the bob to move from A to C is $t_{1}$ and from C to O is $t_{2}$ . The time period of this simple pendulum is