Physics, asked by sprithviraj886, 1 month ago

A spring is stretched by 5 cm by a force 10 N. The time period of the oscillations when a mass of 2 kg is suspended by it is:
(1) 6.28 s
(2) 3.14 s
(3) 0.628 s
(4) 0.0628 s​

Answers

Answered by harshatha16200
2

Answer:

The force constant of the spring,

k=  

x

F

=  

0.2

0.5×10

=25N/m

Now, if the mass of 0.25 kg is suspended by the spring, then the period of oscillation,

T=2π  

k

m

 

=2π  

25

0.25

 

=0.628s

Answered by abhi178
2

Given info : A spring is stretched by 5 cm by a force 10 N.

To find : the time period of the oscillations when a mass of 2 kg is suspended by it is..

solution : we know, spring force , F = kx

where k is spring constant.

here F = spring force  = 10 N

and x = elongation of spring = 5cm = 0.05 m

∴ 10 = k × 0.05

⇒ 10 × 20 = k

⇒ k = 200 N/m

now the time period of the oscillations is given by, T=2\pi\sqrt{\frac{m}{k}}

= 2\pi\sqrt{\frac{2kg}{200N/m}}

= \frac{2\pi}{10} sec

= \frac{2\times3.14}{100}  sec

= 0.628 sec

therefore the time period of the oscillations is 0.628 sec. hence the correct option is (3).

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