Question

One end of an ideal spring is fixed to a wall at origin $O$ and axis of spring is parallel to $x-$axis. A block of mass $m=1kg$ is attached to free end of the spring and it is performing $SHM$. Equation of position of the block in coordinate system shown in figure is $x=10+3\sin 10t$. Here, $t$ is in second and $x$ in $cm$. Another block of mass $M=3kg$, moving towards the origin with velocity $30cm/s$ collides with the block performing $SHM$ at $t=0$ and gets stuck to it calculate.
(i) New amplitude of oscillation
(ii) New equation for position of the combined body.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (C)

$3$

From equation $x=10+3sin\left(10t\right)$

So $\omega =10$ and $A=3$

We know $\omega =\frac{k}{m}$

$k={10}^2\times 1=100N/m$

The velocity of mass m at t=0 is $v_1=\omega \times A=10\times 3=30cm/s$

Now from momentum conservation

$(M+m)v=30\times M-v_1\times m$

$4v=30\times 3-30\times 1$

$v=15cm/s$

Now to find the new amplitude of system

Let the new amplitude be $A_1$

By conservation of energy

$\frac{(M+m)v^2}{2}=\frac{k{A}_1^2}{2}$

$A_1=0.03m=3cm$

and new angular velocity of system ${\omega }_1=\sqrt{\frac{k}{M+m}}$

$\omega =\sqrt{\frac{100}{4}}=5rad/s$

So the equation is $x^{\prime }=10-3sin5t$

$x=3$