Question:
Two oscillating simple pendulums with time periods \(T\) and \(\frac{5T}{4}\) are in phase at a given time. They are again in phase after an elapse of time
Two oscillating simple pendulums with time periods \(T\) and \(\frac{5T}{4}\) are in phase at a given time. They are again in phase after an elapse of time
Show Hint
For phase matching, take LCM of time periods.
Updated On: May 8, 2026
- \(4T\)
- \(3T\)
- \(6T\)
- \(5T\)
- \(8T\)
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Concept:
Two oscillations come in phase again when their phase difference becomes multiple of \(2\pi\), i.e., time equals LCM of periods.
Step 1: Given periods. \[ T_1 = T, \quad T_2 = \frac{5T}{4} \]
Step 2: Find common time. We need smallest \(t\) such that: \[ \frac{t}{T} = n, \quad \frac{t}{5T/4} = m \]
Step 3: Solve using LCM. LCM of \(T\) and \(5T/4\): \[ = 5T \]
Step 4: Conclusion. \[ \boxed{5T} \]
Step 1: Given periods. \[ T_1 = T, \quad T_2 = \frac{5T}{4} \]
Step 2: Find common time. We need smallest \(t\) such that: \[ \frac{t}{T} = n, \quad \frac{t}{5T/4} = m \]
Step 3: Solve using LCM. LCM of \(T\) and \(5T/4\): \[ = 5T \]
Step 4: Conclusion. \[ \boxed{5T} \]
Was this answer helpful?
0
0
Top KEAM Physics Questions
- A body oscillates with SHM according to the equation (in SI units), $x = 5 cos \left(2\pi t +\frac{\pi}{4}\right) .$ Its instantaneous displacement at $t = 1$ second is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of
- Two capillary tubes A and B of diameter $1\, mm$ and $2\, mm$ respectively are dipped vertically in a liquid. If the capillary rise in A is 6 cm, then the capillary rise in B is
- In the circuit given in figure, 1 and 2 are ammeters. Just after key K is pressed to complete the circuit, the reading will be
- A plate has a length $ 5\pm 0.1\text{ }cm $ and breadth $ 2\pm 0.01\text{ }cm $ . Then the area of the plate is:
View More Questions
Top KEAM Pendulums Questions
- Time periods of pendulums \( A \) and \( B \) are \( T \) and \( \frac{5T}{2} \). If they start executing S.H.M. at the same time from the mean position, the phase difference between them after the bigger pendulum has completed one oscillation is
- A particle executes SHM with time period T. If acceleration is doubled keeping amplitudes constant, new time period is
- The frequency of a simple pendulum is f. If its length is increased by four times, its frequency will become
- If the period of a simple pendulum of length \(l\) is 5 seconds, then the period of the pendulum of length \(\frac{l}{4}\) is
- For smaller angular displacement, the period of a simple pendulum depends on
View More Questions
Top KEAM Questions
- i.$\quad$ They help in respiration ii.$\quad$ They help in cell wall formation iii.$\quad$ They help in DNA replication iv. $\quad$They increase surface area of plasma membrane Which of the following prokaryotic structures has all the above roles?
- A body oscillates with SHM according to the equation (in SI units), $x = 5 cos \left(2\pi t +\frac{\pi}{4}\right) .$ Its instantaneous displacement at $t = 1$ second is
- The pH of a solution obtained by mixing 60 mL of 0.1 M BaOH solution at 40m of 0.15m HCI solution is
Kepler's second law (law of areas) of planetary motion leads to law of conservation of
- If $\int e^{2x}f' \left(x\right)dx =g \left(x\right)$, then $ \int\left(e^{2x}f\left(x\right) + e^{2x} f' \left(x\right)\right)dx =$
View More Questions