Question

Block A of maa m, connected with spring of spring constant $K_1$ is executting SHM for which $x=xbloc{k}_0+a_{1}sin\omega t.$ block B of same mass connected with spring of spring connected $K_2$ is executting SHM for $x=x_0-a_{2}sin\omega t$ block B collides with block A elastically at time t=0, then displacement time equation of block A will be

• A. $x_0-a_2\sqrt{\frac{m}{K_1}}sin{\omega }_{1}t$
• B. $x_0-a_1{\omega }_1\sqrt{\frac{m}{K_1}}sin{\omega }_{1}t$
• C. $x_0-a_2{\omega }_2\sqrt{\frac{m}{K_1}}sin{\omega }_{1}t$
• D. $x_0-a_1{\omega }_2\sqrt{\frac{m}{K_1}}sin{\omega }_{1}t$

Answer 1

Answer: (A)

$x_0-a_2\sqrt{\frac{m}{K_1}}sin{\omega }_{1}t$

(A) $x=x_{{block}_0}+a_{1}sinwt.\left(A\right)\left(k_1\right)$

(B) $x=x_0-a_{2}sinwt\left(B\right)\left(k_2\right)$ $t=0$

displacement time equation of block

$A=?$

Spring constant of $A=k_1$

mass of the $A=m$

$x=x_{{block}_0}+a_{1}sinwt$ for block $A$

displacement $=x_0-a_2\sqrt{\frac{m}{k_1}}\sin w_{1}t$