Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Two simple pendulums whose lengths are 121 cm and 169 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, wi

Two simple pendulums whose lengths are 121 cm and 169 cm are suspended side by side. Their bobs are pulled together and then released. After how many minimum oscillations of the longer pendulum, will the two be in phase again.

Option: 1

5.5


Option: 2

10.123


Option: 3

13.678


Option: 4

16.780


Answers (1)

best_answer

Let  T_1 and T_2 are the time period of the two pendulums 

\mathrm{T}_1=2 \pi \sqrt{\frac{121}{\mathrm{~g}}} \text { and } \mathrm{T}_2 \pi \sqrt{\frac{169}{\mathrm{~g}}}\left(\mathrm{~T}_1<\mathrm{T}_2 \text { because } \mathrm{l}_1<\mathrm{l}_2\right)

Let at t=0, they start swinging together. Since their time periods are different, the swinging will not be in unison always. 

Only when the number of complete oscillations differs by an integer, the two pendulums will again begin to swing together.

Let longer length pendulum complete \mathrm{n} oscillation and shorter length pendulum complete (n+1) oscillation for the incision swinging, then (n+1) T_1=n T_2

(n+1) \times 2 \pi \sqrt{\frac{121}{g}}=n \times 2 \pi \sqrt{\frac{169}{g}}

Or, \mathrm{n}=5.5

Posted by

Info Expert 30

View full answer

JEE Main high-scoring chapters and topics

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

Similar Questions

Latest Question