Question

The bob of a perpendicular oscillation 30 times per minute.find its time period and frequency

Answer 1

see, we know that frequency is no. of oscillations made by bob of a pendulum in one second.So,

Frequency = 30x60 = 1800 Hz per sec.

now, Time period = 1/frequency

so, Time period is= 1/1800 s.

Hope, it helped you.

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Answer 2

The time period and frequency is 2 seconds and 0.5 Hz.

Explanation:

It is given that, the bob of a perpendicular oscillation 30 times per minute. The frequency of an object is given by the total number of oscillations per second. Here, oscillation of per minute is 30. The frequency of a perpendicular oscillation is :

$f=\dfrac{30}{60}=0.5\ Hz$

The reciprocal of frequency is called its time period. So,

$T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.5}\\\\T=2\ s$

So, the time period and frequency is 2 seconds and 0.5 Hz.

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