Please wait...
Contact Us
Contact
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

Reply fast plzz??

qsnImg
Asked by SanskarAgarwal86 26th February 2019, 4:17 PM
Answered by Expert
Answer:
Force constant k of spring is obtained from the relation,  F = kx , where F is force applied to spring and x is extension
 
hence k = F/x = (0.2×9.8)/(4.9×10-2 ) = 40 N/m
 
Period of oscillation T = begin mathsize 12px style 2 straight pi cross times square root of straight m over straight k end root space equals space 2 straight pi cross times square root of fraction numerator 0.4 over denominator 40 end fraction end root space equals space 0.628 space straight s end style
Energy stored in spring when it is extended  (1/2)kx2
 
Change in energy when it is extended to 9.8 cm from 4.9 cm = (1/2)×40×[ 9.8×9.8 - 4.9×4.9 ]×10-4 = 0.144 J
Answered by Expert 26th February 2019, 4:54 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp