Question

If the oscillation is 20 and time taken is 60seconds ,what is the time period

Answer 1

Answer:

Given:

Number of oscillations = 20

Time taken = 60 seconds

To Find:

Time period of the oscillatory object

Definitions:

Time period is the time taken by an oscillatory object to complete one complete oscillation.

Concept:

Since we have been given Total number of oscillations in total time, we can use "Unitary Method" to find out the time for one oscillation.

Calculation:

20 oscillations in 60 seconds

=> 1 oscillation in 60/20 = 3 seconds

So time period (as per definition) is 3 seconds.

Final Answer:

3 seconds.

Answer 2

Question :-

If the oscillation is 20 and time taken is 60seconds ,what is the time period.?

Answer :-

$\sf{3 seconds}$

To Find :-

$\sf{ time period of 1 oscillation}$

Solution :-

Given that ,

• Number of oscillations = 20

• time taken = 60 seconds

We used unitary method ,

•.• According to the question,

° 20 oscillations taken time = 60 s

hence,

1 oscillation taken time = 60/20 = 3

therefore,

time period is 3 seconds in 1 oscillation.

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Some important Formula :-

$\boxed{ \star } \: \: \: \sf{\rho \: = \frac{m}{v} } \\ \rightarrow \: \rho \: \sf{is \: density }\: \\ \\ \boxed{ \star} \: \sf{\:force(f) \: = m \: a }\: \\ \rightarrow \sf{a \: is \: acceleration \: } \: \\ \rightarrow \sf{m \: be \: the \: mass \: of \: object} \\ \\ \boxed{\star} \: \: \sf{ \: p = \frac{w}{t} } \\ \\ \rightarrow \sf{p \: is \: power \: } \\ \rightarrow \sf{w \: is \: work \: done \: } \\ \\ \boxed{ \star }\: \sf{pressure(p) = \frac{force(f)}{area(a)} } \\ \\ \boxed{\star} \: \sf{kinatic \: energy \: = \frac{1}{2} \: m \: {v}^{2} }$

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