Question

Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter13). Give the signs of position, velocity and acceleration variables of the particle at \(t\)= \(0.3 \,s, 1.2 \,s, – 1.2 \,s.\)

Answer 1

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: 
\(a\) =\(– \omega^2 \times \omega \rightarrow angular \,frequency\) ....(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.