Question

The displacement-time graph of a particle executing SHM is shown in figure. Which of the following statement is/are true ?

A. The focrce is zero at `t=(3T)/(4)`
B. The acceleration is maximum at `t=(4T)/(4)`
C. The velocity is maximum at `t=(T)/(4)`
D. The PE is equal to KE of oscillation at `t=(T)/(2)`

Answer 1

Consider the diagram

From the given diagram, it is clear that
(a) A `t t=(3T)/(4)` the displacement of the paritcle is zera. Hence, the particle execution SHM will be at mean position i.e., x=0. So, acceleration is zero and force also zero.
(b) A `t t=(4T)/(4)` displacement is maximum i.e., extreme position, so acceleration is maximum.
(c) A `t t=(T)/(4)` corresponds to mean postion, so velocitly will be maximum at this position.
(d) A `t t=(2T)/(4)=(T)/(2)` corresponds to extreme position, so `KE=0"and"PE=",maximum"`.