Answer:
Refer to Fig 10 (NCT). 2, where A and 5 represent the two
extreme positions of a S.H.M. For velocity, the direction from A to B is taken
to be positive. The acceleration and the force .a long AP are taken as positive;
and along BP are taken as negative.
(a)
At the end A, the particle executing S.H.M. is momentarily at rest being its
extreme position of motion. Therefore, its velocity is zero; acceleration is
+ve because it is directed along AP. Force is also +ive since the force is
directed along AP, i.e., +ive direction.
(b)
At the end B, velocity is zero. Here, acceleration and force are negative as
they are directed along BP, i.e., along negative direction.
(c)
At the midpoint of AB going towards A, the particle is at its mean position P,
with a tendency to move along PA, i.e., -ve direction. Hence velocity is -ve.
Both, acceleration and force are.wro.
(d)
At 2 cm away from B going towards A, the particle is at Q, with a tendency to
move along QP, which is negative direction. Therefore, velocity, acceleration and
force, all are negative.
(e)
At 3 cm away from A going towards B, the particle is at R, with a tendency to
move along RP, which is positive direction. Here, velocity, acceleration and
force, all are positive.
(f)
At 4 cm away from A going towards A, the particle is at S, with a tendency to
move along SA, which is negative direction for velocity. Therefore, velocity is
negative but acceleration is directed towards mean position, i.e., along SP,
hence +ve. Similarly, force is also +ve.
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