Question

When a body of mass 1.0 kg is suspended from a certain light spring hanging vertically, its length increases by 5 cm. By suspending the 2.0 kg block to the spring and if the block is pulled through 10 cm and released the maximum velocity in it in m/s is :( Acceleration due to gravity\[ = \,10\,m/{s^2}\])
A. 0.5
B. 1
C. 2
D. 4

Answer 1

Hint: Maximum velocity is product of amplitude and angular frequency, amplitude is considered as the maximum displacement from equilibrium position and velocity is maximum when an object passes through its equilibrium position.

Formula used :
\[1.{\rm{Spring \,force, }}F = mg = kx\,\,\]
where, K = Spring constant and x = Elongation in spring.
\[2.\,{\rm{Maximum}}\,{\rm{velocity,}}\,{v_{\max }} = A\omega \]
Where, A = Amplitude and $\omega$ = angular frequency
\[3.\,\,{\rm{Angular}}\,{\rm{Frequency,}}\,\omega = \sqrt {\dfrac{k}{m}} \]
Where, m = mass of the system.

Complete step by step solution:
Given here is a body of mass 1.0 kg suspended from a spring and as result spring is elongated by 5 cm, we have to find the maximum velocity when a mass 2.0 kg is suspended from the spring and block is released after pulling the spring by 10 cm. For mass \[{m_1} = \,1.0\,kg\] and elongation \[{x_1} = 5\,cm = \dfrac{5}{{100}}m\] using Hooke’s law spring force is given as,
\[{\rm{ }}{F_1} = {m_1}g = k{x_1}\,\,\]

Substituting value of $m_1$ and $x_1$ in above equation we get,
\[{\rm{ }}1.0 \times 10 = k \times \dfrac{5}{{100}} \\
\Rightarrow k = \dfrac{{10 \times 100}}{5}\, = 200\,N/m\]
For mass, \[{m_2} = \,2.0\,kg\] and elongation \[{x_2} = 10\,cm = \dfrac{{10}}{{100}}m\].
As the velocity is maximum at \[{x_2} = 10\,cm = \dfrac{{10}}{{100}}m\] therefore this displacement will be considered as amplitude (A) of oscillation i.e.
\[A = {x_2} = \dfrac{{10}}{{100}}m\] and,
Maximum velocity, \[{v_{\max }} = A\omega \]
where \[\omega = {\rm{Angular}}\,{\rm{frequency}}\] and it is given by
\[\omega = \sqrt {\dfrac{k}{m}} \]
Where, K = Spring constant and m =mass of the system.

Therefore, maximum velocity can be written as,
\[{v_{\max }} = A\omega = A\sqrt {\dfrac{k}{m}} \]
Substituting \[A = \dfrac{{10}}{{100}}m\], \[k = 200\,N/m\] and m= 2.0 kg in above equation we get,
\[{v_{\max }} = \dfrac{{10}}{{100}}\sqrt {\dfrac{{200}}{{2.0}}} \\
\Rightarrow {v_{\max }} = 0.1 \times 10 = 1\,m/s\]
Hence, maximum velocity when spring is released from 10 cm is 1 m/s.

Therefore, option B is the correct answer.

Note: Maximum velocity of an oscillating body occurs when body passes through is equilibrium, for a pendulum it would occur when the bob is closest to ground, and for spring mass system it would occur when spring is neither stretched or compressed.