A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How…

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Question “A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How…”

Answer

This problem can be solved using two concepts: displacement and time for simple harmonic motion.

Simple harmonic motion is taking place on the machine part.

To find the time it takes for displacement of simple harmonic motion in the interval

to , use the expression for displacement of simple harmonic motion.

Simple harmonic motion is simply velocity, which oscillates just like displacement.

Simple harmonic motion refers to periodic motion in the which the restoring force is proportional to the displacement.

The time period refers to the time it takes for one oscillation.

The simple harmonic motion of displacement is expressed as,

represents the displacement.

indicates the amplitude.

determines the frequency.

specifies the time.

The maximum displacement from equilibrium is called amplitude.

The simple harmonic motion of displacement is expressed as,

The value of

at the initial position is zero.

Substitute HTMLmedia_tag_9$ to HTMLmedia_tag_10$ or HTMLmedia_tag_11$ to

.

indicates the time

.

The above expression is an example.

The simple harmonic motion of displacement is expressed as,

The final position of

has

.

Substitute

to

and

to

.

is for

.

this is when #media_tag_28$.

Expression of the time it took to replace in the

to

interval is

refers to the time it took for the displacement between

and

.

Substitute HTMLmedia_tag_35$ to

or HTMLmedia_tag_37$ to #media_tag_38$.

Substitute $50$ for $41$.

The time required to replace the

with

was

.

Ans:

The time it took to replace the

with

was

.

Conclusion

Above is the solution for “A machine part is undergoing SHM with a frequency of 5.00Hz and amplitude 1.80cm . How…“. We hope that you find a good answer and gain the knowledge about this topic of science.