14.16 Answer the following questions :

(a) Time period of a particle in SHM depends on the force constant k and mass m of the particle

T = 2 $π \sqrt{\frac{m}{k}} .$ A simple pendulum executes SHM approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?

(b) The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that T is greater than 2 $π \sqrt{\frac{l}{g}}$ . Think of a qualitative argument to appreciate this result.

(c) A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall ?

(d) What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely falling under gravity?

2 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

(a) The time period of a simple pendulum, T = 2 $π \sqrt{\frac{m}{k}}$

For a simple pendulum, k is expressed in terms of mass, m as : k $\propto m$ or $\frac{m}{k}$ = constant

Hence, the time period of a simple pendulum is independent of the mass of the bob. In the case of a simple pendulum, the restoring force acting on bob is given as F = -mg $\sin ? θ$ , where

F = restoring force

m = mass of the bob

g = acceleration due to gravity

$θ = a n g l e o f d i s p l a c e m e n t$

(b) For small $θ,$ sin $θ$ $~ θ$ . For larger $θ,$ sin $θ$ is greater than $θ$ . This decreases the effective value of g.

Hence the time period increase as : T = 2 $π \sqrt{\frac{l}{g}}$ , where l is the length of

...more

(a) The time period of a simple pendulum, T = 2 $π \sqrt{\frac{m}{k}}$

For a simple pendulum, k is expressed in terms of mass, m as : k $\propto m$ or $\frac{m}{k}$ = constant

Hence, the time period of a simple pendulum is independent of the mass of the bob. In the case of a simple pendulum, the restoring force acting on bob is given as F = -mg $\sin ? θ$ , where

F = restoring force

m = mass of the bob

g = acceleration due to gravity

$θ = a n g l e o f d i s p l a c e m e n t$

(b) For small $θ,$ sin $θ$ $~ θ$ . For larger $θ,$ sin $θ$ is greater than $θ$ . This decreases the effective value of g.

Hence the time period increase as : T = 2 $π \sqrt{\frac{l}{g}}$ , where l is the length of the simple pendulum.

(c) The time shown by the wristwatch of a man falling from the top of the tower is not affected by the fall. Wrist watch works on spring action, independent of acceleration due to gravity.

(d) During free fall under gravity, the acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

less

(a) The time period of a simple pendulum, T = 2 <math><mi>π</mi><msqrt><mrow><mfrac><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>k</mi></mrow></mrow></mfrac></mrow></msqrt></math>For a simple pendulum, k is expressed in terms of mass, m as : k <math><mo>∝</mo><mi>m</mi></math> or <math><mfrac><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>k</mi></mrow></mrow></mfrac></math> = constantHence, the time period of a simple pendulum is independent of the mass of the bob. In the case of a simple pendulum, the restoring force acting on bob is given as F = -mg <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>? </mo><mrow><mi>θ</mi></mrow></mrow></math> , whereF = restoring forcem = mass of the bobg = acceleration due to gravity<math><mi>θ</mi><mo>=</mo><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">g</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">f</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">d</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">p</mi><mi mathvariant="normal">l</mi><mi mathvariant="normal">a</mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">m</mi><mi mathvariant="normal">e</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">t</mi></math>  (b) For small <math><mi>θ</mi><mo>, </mo><mi></mi></math> sin <math><mi></mi><mi>θ</mi></math> <math><mo>~</mo><mi></mi><mi>θ</mi></math> . For larger <math><mi>θ</mi><mo>, </mo></math> sin <math><mi></mi><mi>θ</mi></math> is greater than <math><mi>θ</mi></math> . This decreases the effective value of g.Hence the time period increase as : T = 2 <math><mi>π</mi><msqrt><mrow><mfrac><mrow><mrow><mi>l</mi></mrow></mrow><mrow><mrow><mi>g</mi></mrow></mrow></mfrac></mrow></msqrt></math> , where l is the length of the simple pendulum.  (c) The time shown by the wristwatch of a man falling from the top of the tower is not affected by the fall. Wrist watch works on spring action, independent of acceleration due to gravity.  (d) During free fall under gravity, the acceleration is zero. Hence the frequency of oscillation of this simple pendulum is zero.

0 Upvotes 0 Downvotes

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question