Question

A student performs experiment with simple pendulum and measures time for 10 vibrations. If he measures a time for 100 vibrations, the relative error in measurements of time period will be reduced by? In this question I did not understand this step --> now % of the error reduced is given by: 10−1/10×100 =90%. Please explain this step.

Answer 1

Hello.. Here is your answer

% error for 10 vibrations is 10 %

% error for 100 vibrations is 1 %

% of the error reduced is given by =[First error − second error/First error] ×100

=[10−1/10×100] =90%

Regards

Answer 2

Answer:

Let the relative error for 10 vibrations be

⇒K =Δ$\frac{T}{10}$

Now Relative error for 100 vibrations be

K' = Δ$\frac{T}{100}$ = $\frac{K}{10}$

⇒ Reduction in relative error is

$\frac{K' - K}{K}$×100%

= $\frac{9}{10}$×100%

= 90%

Frequency and period are inversely related . The pendulum with the smallest period will have the highest frequency of vibration . A longer pendulum has a higher period ; a shorter pendulum has a smaller period.

Thus, the pendulum with shorter string will have a higher frequency of vibration.

A simple pendulum is another mechanical system that moves in an oscillatory motion.

Learn more about simple pendulum vibration :

https://brainly.in/question/12054286

https://brainly.in/question/1645741

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