Question

a mosquito flaps its wings at a rate of 500 per second(vibrations) calculate the time period of the vibrations.:×)

Answer 1

Hey MATE!

Time period is actually the time when one flap is completed by the mosquito.

500 flaps --------- 1sec

1 falp --------- 1/500 sec = 0.002 sec Answer

Hope it helps

Hakuna Matata :))

Answer 2

Given : A mosquito flaps its wings at a rate of 500 per second (vibrations).

　

To find : What is its time period?

　

Using formula :

1) Frequency = Number of times the pendulum oscillates/Total number of time taken.

　

2) Time period = Number of oscillation per second/Frequency.

　

Calculations :

→ Frequency = 500/1

→ Frequency = 500 hertz

　

→ Time period = 1/500

→ Time period = 0.002 seconds

　

Therefore,

1. Frequency = 500 hertz.
2. Time period = 0.002 seconds.