Question

Question 3.20 Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.


(Fig: 3.23)

Chapter Motion In A Straight Line Page 58

Answer 1

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From the figure ,
We know simple harmonic motion satisfied condition.
a = -w²x
Where A , w is positive constant
(A) at t = 0.3
X is negative
Slope of graph at t = 0.3 negative so, velocity < 0 .
a will be positive because displacement is negative { a=-w²x}

(B) at t = 1.2
X is positive { shown in figure }
slope of displacement - time graph is positive at t = 1.2 so, velocity will be positive .
acceleration will be negative because displacement is positive .
(C) at t = -1.2
Displacement is negative.
Slope of graph at t= -1.2 is positive so, velocity at t = -1.2 is positive .
Acceleration will be positive because displacement is negative.

Answer 2

Answer:

From the figure ,

We know simple harmonic motion satisfied condition.

a = -w²x

Where A , w is positive constant

(A) at t = 0.3

X is negative

Slope of graph at t = 0.3 negative so, velocity < 0 .

a will be positive because displacement is negative { a=-w²x}

(B) at t = 1.2

X is positive { shown in figure }

slope of displacement - time graph is positive at t = 1.2 so, velocity will be positive .

acceleration will be negative because displacement is positive .

(C) at t = -1.2

Displacement is negative.

Slope of graph at t= -1.2 is positive so, velocity at t = -1.2 is positive .

Acceleration will be positive because displacement is negative.

Explanation: