Question

The weight of mass $1$ kg attached to a spring with force constant 30 N/m is able to oscillate in a horizontal steel rod. The initial displacement from equilibrium position is $30$ cm. Given: $g=10m/s^2$ and coefficient of friction is $\mu =0.1$ If, before stopping completely, then the weight swings $N$ times, $N$ is:

• A. 8
• B. 6
• C. 5
• D. 9

Answer 1

The correct option is A 8

The frictional force$=\mu mg$
The weight goes over from initial state of maximum deflection to a position of equilibrium characterized by amplitude $A_0$ to another similar state with amplitude $A_1$.
By conservation of energy,
$\frac{1}{2}K{A}_0^2-\mu mg{A}_0=\frac{1}{2}k{A}_1^2+\mu mg{A}_1$
$\Rightarrow A_1=A_0-\frac{2\mu mg}{k}$
This will be true for subsequent oscillations. The amplitude form an arithmetic progression with general term being $A_n=A_0$$-\left(\frac{2\mu mg}{k}\right).n$
The weight will stop when the amplitude become zero.
i.e.,if $A_n=0\Rightarrow n=\frac{k{A}_0}{2\mu mg}$
Since all the amplitude except the initial one are passed through twice the number of swings,
$\Rightarrow N=2n-1=\frac{k{A}_0}{\mu mg}-1=\left(\frac{30\times 0.3}{0.1\times 1\times 10}\right)-1=9-1=8$