A body of mass \(\mathrm{I} \mathrm{kg}\) is suspended from a spring of negligible mass. Another body of…
A body of mass \(\mathrm{I} \mathrm{kg}\) is suspended from a spring of negligible mass. Another body of mass \(500 \mathrm{~g}\) moving vertically upwards hits the suspended body with a velocity of \(3 \mathrm{~ms}^{-1}\) and gets embedded in it. If the frequency of oscillation of the system of the two bodies after collision is \(\frac{10}{\pi} \mathrm{Hz}\), the amplitude of the motion and the spring constant are respectively,
\(5 \mathrm{~cm}, 300 \mathrm{Nm}^{-1}\)
\(10 \mathrm{~cm}, 300 \mathrm{Nm}^{-1}\)
\(10 \mathrm{~cm}, 600 \mathrm{Nm}^{-1}\)
\(5 \mathrm{~cm}, 600 \mathrm{Nm}^{-1}\)
Solution
Given, mass of body, \(M=1 \mathrm{~kg}\), frequency of oscillation, \(f=\frac{10}{\pi} \mathrm{Hz}\),
mass of hitting body, \(m=0.5 \mathrm{~kg}\) and speed of body, \(v=3 \mathrm{~ms}^{-1}\)
As frequency of oscillation of spring mass system,
\(f=\frac{1}{2 \pi} \sqrt{\frac{k}{M_{\text {body }}}}\)
Since, hitting body get embedded to the initial mass, so mass of the body changed to
\(M_{\text {body }}=M+m=1+0.5=1.5 \mathrm{~kg}\)
Spring constant, \(\quad k=(2 \pi /)^2 M_{\text {body }}\)
\(=\left(2 \pi \times \frac{10}{\pi}\right)^2(1.5)=600 \mathrm{Nm}^{-1}\)
Now, velocity of body after collision,
\(\begin{aligned}
& (M+m) v^{\prime}=m v \Rightarrow(1+1 / 2) v^{\prime}=1 / 2 \times 3 \\
& v^{\prime}=1 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
The maximum amplitude of the sping-mass system is \(v^{\prime}=A \omega \Rightarrow A=\frac{v^{\prime}}{2 \pi f}=\frac{1}{20}=0.05 \mathrm{~m}=5 \mathrm{~cm}\)
Hence, the correct answer is (d).