Question

Figure give the x-t plots of a particle executing one-dimensional simple harmonic motion

Match the Column I with Column II

	Column I Time		Column II Signs of position (x), velocity (v) and acceleration (1)
(A)	At t = -1.2 s	(p)	x < 0, v < 0, a > 0
(B)	At t = -0.3 s	(q)	x > 0, v > 0, a < 0
(C)	At t = 0.3 s	(r)	x > 0, v < 0, a < 0
(D)	At t = 1.2 s	(s)	x < 0, v > 0, a > 0

• A. A – s, B – r, C – p, D = q
• B. A – p, B – q, C – r, D = s
• C. A – q, B – p, C – s, D = r
• D. A – r, B – s, C – q, D = p

Answer 1

Answer: (A)

A – s, B – r, C – p, D = q

Answer 2

In SHM, accelerations , a = $- \omega^{2}x$

Where $\omega$ (i.e., angular frequency) is constant.

At $t = - 1.2s,x < 0$

The slope of x –t is positive, hence v is positive.

Since $a = - \omega^{2}x,$ hence a is positive

$\therefore$At t = - 1.2 s, x < 0 , v > 0, a > 0

A – s

At $t = - 0.3s,x > 0$

The slope of x-t is negative hence v is negative.

Since a $= - \omega^{2}x,$ hence a < 0

$\therefore Att = 0.3s,x > 0,v < 0,a < 0$ B –r

At $t = 0.3x,x < 0$

The slope of x – t is negative hence v is negative.

Since $a = \omega^{2}x$hence a > 0

C – p

At $t = 1.2s,x > 0$

The slope of x – t is positive, hence v is positive,

Since a $= \omega^{2}x$hence a < 0

$\therefore$At t $= 1.2s,$ x > 0, v > 0, a < 0

D- q