Question

if a pendulum vibrates with frequency 0.5Hz then what its length?Explain

Answer 1

$frequency \: of \: pendulum \\ = \frac{1}{2\pi} \sqrt{ \frac{g}{l} } \\ where \: g \: is \: gravity \: and \: l \: is \: the \: length \: of \: pendulum \: \\ here \\ 0.5 = \frac{1}{2\pi} \sqrt{ \frac{g}{l} } \\ taking \: g \: as \: 9.8 \\ 0.5 = \frac{1}{2\pi} \sqrt{ \frac{9.8}{l} } \\ 0.5 \times 2\pi = \sqrt{ \frac{9.8}{l} } \\ 1 \times \frac{22}{7} = \sqrt{ \frac{9.8}{l} } \\ \frac{22}{7} \times \frac{22}{7} = \frac{9.8}{l} \\ l = \frac{9.8 \times 7 \times 7}{22 \times 22} \\ l = 0.9921m$

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Answer 2

Concept:

The frequency of a pendulum controls how frequently it swings back and forth over a predetermined period of time.

It is mathematically expressed as- f = 1/2π√g/l

Given:

Frequency of pendulum = 0.5 Hz.

Find:

We need to determine the length of the pendulum

Solution:

The frequency of a pendulum controls how frequently it swings back and forth over a predetermined period of time.

The frequency of a pendulum is given by the equation, f = 1/2π√g/l

here, g is the gravity while l is the length of the pendulum

We have frequency, f = 0.5 Hz

We know, g = 9.8 m/s² and π = 22/7

Therefore, the equation for the frequency of a pendulum becomes-

0.5 = 1/2π √9.8/L

0.5 × 2π = √9.8/L

0.5 × 2 × 22/7 = √9.8/L

22/7 = √9.8/L

Taking square on both sides

(22/7 × 22/7) = 9.8/L

L = 9.8 × 7/22 × 7/22

L = 0.9921 m is the length of the pendulum

Thus, the length of the pendulum is 0.9921m.

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