Solve
Guides
0
Question
The displacement-time graph of a particle executing SHM is shown in the figure. Then
- the acceleration is maximum at t=T
- the force is zero at t=3T/4
- the potential energy equals the oscillation at t=T/2
- the velocity is maximum at t=T/2
Open in App
Solution
Verified by Toppr
From the graph, velocity is maximum at t=0,T
The acceleration ,d2ydt2 is maximum at t=T
Force is zero at t=T/2,3T/4
As the interval of max and min displacement is T2, so the potential energy equals the oscillation at T/2.
Was this answer helpful?
0
Similar Questions
Q1
The displacement time graph of a particle executing S.H.M. is given in figure : (sketch is schematic and not to scale)
Which of the following statements is/are true for this motion?
(a)The force is zero at t=3T4
(b)The acceleration is maximum at t=T
(c)The speed is maximum at t=T4
(d)The P.E. is equal to K.E. of the oscillation at t=T2
Which of the following statements is/are true for this motion?(a)The force is zero at t=3T4
(b)The acceleration is maximum at t=T
(c)The speed is maximum at t=T4
(d)The P.E. is equal to K.E. of the oscillation at t=T2
View Solution
View Solution
Q3
Displacement time graph of particle performing SHM is shown in figure. Assume that mean position is at x=0, then kinetic energy and potential energy will be equal at:
[T is time period and A is amplitude]
[T is time period and A is amplitude]
View Solution
Q4
Acceleration-displacement graph of a particle executing SHM is as shown in the figure. The time period of oscillation is (in sec)
View Solution
Q5
Acceleration-time graph of a particle moving in a straight line is as shown in the figure. At time t=0, velocity of the particle is zero. Find velocity of the particle at t=14s:
View Solution