First slide

Simple hormonic motion

Question

A block is suspended by a spring and the elongation produced in the spring is found to be $x_{0}$ . Then the block is further pulled downward by a distance $x_{0}$ and released. Then acceleration of the block in its lower most position is

Moderate

A

g
B

g/2
C

2g
D

zero

Solution

After release, the block will execute vertical simple harmonic motion about is equilibrium position with amplitude $(2 x_{0} - x_{0})$ i.e. $x_{0}$ and angular frequency $ω$
Here $ω = \sqrt{\frac{k}{m}}$ and $x_{0} = \frac{m g}{k}$
$∴$ Acceleration at the lower extreme position $= ω^{2} x_{0} = {(\frac{K}{m})}^{} \times \frac{m g}{k} = g$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App