#### Question

Figure gives the *x-t *plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at *t *= 0.3 s, 1.2 s, – 1.2 s.

#### Solution 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* = – 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.

#### Solution 2

In x-t graph of Fig. showing simple harmonic motion of a particle, the signs of position, velocity and acceleration are as given below.

In S.H.M., acceleration, a – x or a = – kx.

(i) At t = 0.3 s, x < 0 i.e., x is in -ve direction. Moreover, as x is becoming more negative with time, it shows that v is also – ve (i.e., v < 0). However, a = -kx will be +ve (a > 0).

(ii) At t = 1.2 s, x > 0, v > 0 and a < 0.

(iii) At t = -1.2 s, x < 0, but here on increasing the time t, value of x becomes less negative.

It means that v is +ve (i.e., v > 0). Again a = – kx will be positive (i.e., a > 0).