Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Need explanation for: - Oscillations and Waves - JEE Main-6

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with a acceleration a, then the time period is given by T = 2\pi\sqrt{\frac{l}{g'}}, where g' is equal to 

  • Option 1)

    g

  • Option 2)

    g - a

  • Option 3)

    g + a

  • Option 4)

    \sqrt{g^{2} + a^{2}}

 

Answers (1)

best_answer

g' = \sqrt{g^{2}+a^{2}}

 

Time period of simple pendulum accelerating horizontally -

T= 2\pi \sqrt{\frac{l}{g^{2}+a^{2}}}

- wherein

l= length of pendulum

g= acceleration due to gravity.

a= acceleration of pendulum.

 

 

 


Option 1)

g

This is incorrect.

Option 2)

g - a

This is incorrect.

Option 3)

g + a

This is incorrect.

Option 4)

\sqrt{g^{2} + a^{2}}

This is correct.

Posted by

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JAC Chandigarh Counselling 2025: Round 1 seat allotment result out for BTech admissions
anam-khan

JoSAA round 5 seat allotment result 2025 out on josaa.nic.in
anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today

Latest Question