Physics - SHM Question with Solution | TestHub

Initially, the block of mass $m=1$ kg oscillates with amplitude $A=0.1$ m and spring constant $k=100$ N/m. The maximum velocity of the block at the equilibrium position is $v_{max}=A\omega =A\sqrt{\frac{k}{m}}$.$$v_{max}=(0.1)\sqrt{\frac{100}{1}}=(0.1)(10)=1\text{ m/s}$$

When the second block of mass $m^{\prime }=3$ kg is gently placed on it and sticks, momentum is conserved just before and after the collision.$$m{v}_{max}=(m+m^{\prime })v_{max}^{\prime }$$ $$(1)(1)=(1+3)v_{max}^{\prime }\Rightarrow 1=4v_{max}^{\prime }\Rightarrow v_{max}^{\prime }=0.25\text{ m/s}$$ The new system has a total mass $M_{total}=m+m^{\prime }=4$ kg. The spring constant remains $k=100$ N/m. The new angular frequency is ${\omega }^{\prime }=\sqrt{\frac{k}{M_{total}}}=\sqrt{\frac{100}{4}}=5$ rad/s. The new amplitude $A^{\prime }$ is related to the new maximum velocity by $v_{max}^{\prime }=A^{\prime }{\omega }^{\prime }$.$$0.25=A^{\prime }(5)\Rightarrow A^{\prime }=\frac{0.25}{5}=0.05\text{ m}$$ Converting to cm, $A^{\prime }=0.05\times 100=5$ cm.