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
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
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,
=
= sec
= 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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