Question

a mass of 98 kg is vibrating at the end of a spring of force constant 200N/m it's time period is​

Answer 1

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• 1 Newton = 1 Kg × 1 m/s

• Time period = 1 / Frequency

Here , Frequency =

$\frac{2000 \: n /m }{98 \: kg}$

or ,

$\frac{( \: 2000 \: \times \: kg \: \times m / s \:) \: /\: m }{98 \: \times \: kg}$

or ,

$\frac{( \: 2000 \: \times \: kg \: \times 1 / s \:) }{98 \: \times \: kg}$

or ,

$\frac{2000 \times 1 /s}{98}$

or ,

$\frac{2000 \times 1 }{98 \: \times s}$

or ,

$\frac{20.40}{s}$

or , Frequency = 20.40

or , Frequency = 20.40( Because , it makes 20.40 oscillations per second )

Now , Time period - /

= 1 / Frequency

= 1 / 20.40

= 0.05 Seconds ( Approximately )

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Answer 2

Answer:

The time period of oscillation is 4.396 sec .

Explanation:

Given that :

• Mass, m = 98 kg
• Spring constant of spring = 200N/m

To find :

• Time period of oscillation spring.

Solution :

• Let, the body of mass, m is attached to a spring of spring constant, k.
• Let, this body oscillates with a time period, T. The time period of oscillation is given by :

$T = 2\pi \sqrt{ \frac{m}{k} }$

• Putting values in above formula, we get ;

$T = 2\pi \sqrt{ \frac{98}{200} } \\ = 2\pi \sqrt{ \frac{49}{100} } \\ = 2\pi( \frac{7}{10} ) \\ = \frac{7}{5} \pi \\ T = 4.396 \: sec$

• Hence, time period of oscillation is 4.396 sec.