__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 Chapter 14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.__

__Solution:__

Negative, Negative, Positive (at

*t*= 0.3 s)

Positive, Positive, Negative (at

*t*= 1.2 s)

Negative, Positive, Positive (at

*t*= –1.2 s)

For simple harmonic motion (SHM) of a particle, acceleration (

*a*) is given by the relation:

*a*= – ω

^{2}

*x*ω → angular frequency … (i)

*t*= 0.3 s

In this time interval,

*x*is negative. Thus, the slope of the

*x-t*plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.

*t*= 1.2 s

In this time interval,

*x*is positive. Thus, the slope of the

*x*-

*t*plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.

*t*= – 1.2 s

In this time interval,

*x*is negative. Thus, the slope of the

*x*-

*t*plot will also be negative. Since both

*x*and

*t*are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.

## No comments:

## Post a Comment