Question

A particle is executing simple harmonic motion between extreme positions given by $(-1.-2.-3)cm$ and $(1,2,1)cm$. Its amplitude of oscillation is :

• A. $6cm$
• B. $4cm$
• C. $2cm$
• D. $3cm$

Answer 1

Answer: (D)

$3cm$

Let $P(-1,-2,-3)cm$ and $Q(1,2,1)cm$ are the points of extreme positions in S.H.M.

Then $2\times A=d(P,Q)$

$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$

$=\sqrt{(1-(-1){)}^2+(2-(-2){)}^2+(1-(-3){)}^2}$

$=\sqrt{2^2+4^2+4^2}$

$2A=\sqrt{36}$

Amplitude $=\frac{6}{2}=3cm$