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.

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Solution

The given graph is shown below:

From the graph at t=0.3 s, the position is negative and the slope of x-t graph is also negative. Therefore, the sign of both the position and velocity is negative. This means x<0 and v<0 at t=0.3 s.

The acceleration in simple harmonic motion is given as,

a=−kx

The sign of position is negative. Therefore, the sign of acceleration is positive or a>0.

From the graph at t=1.2 s, the position is positive and the slope of x-t graph is also positive. Therefore, the sign of position and velocity is positive. This means x>0 and v>0 at t=1.2 s.

The acceleration in simple harmonic motion is given as,

a=−kx

The sign of position is positive. Therefore, the sign of acceleration is negative or a<0.

From the graph at t=−1.2 s, the position is negative and the time is also negative, therefore the slope of x-t graph is positive. This means x<0 and v>0 at t=−1.2 s.

The acceleration in simple harmonic motion is given as,

a=−kx

The sign of position is negative. Therefore, the sign of acceleration is positive or a>0.

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