Question

1 point
10) The vertical extension in a
light spring by a weight of 1 kg
suspended from the wire is 9.8
cm. The period of oscillation is
*

a) 2pi sec
b) 2 pi sec
c) 2pi/10 sec
d) 200 pi sec​

Answer 1

Answer:

• c) 2π/10  sec

Given :

• The vertical extension in a  light spring by a weight of 1 kg  suspended from the wire is 9.8  cm.

To Find :

• The period of oscillation

Solution :

Length of wire , l = 9.8 cm = 9.8/100 m

Gravity , g = 9.8 m/s²

Now , net force along y-direction is ,

⇒ kx - mg = 0 [ Since net force along y-direction is zero ]

⇒ kx = mg

⇒ k = mg/x

⇒ k = ( 1*9.8 ) / ( 9.8/100 )

⇒ k = 100

$\bf T=2\pi \sqrt{\dfrac{m}{k}}\\\\\implies \bf T=2\pi \sqrt{\dfrac{1}{100}}\\\\\implies \bf T=\dfrac{2\pi}{10} \;sec$

So option (c) is correct