- Home
- Standard 11
- Physics
13.Oscillations
medium
Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. The amplitude of oscillations is
A
$\frac{{{m_1}g}}{K}$
B
$\frac{{{m_2}g}}{K}$
C
$\frac{{({m_1} + {m_2})g}}{K}$
D
$\frac{{({m_1} - {m_2})g}}{K}$
Solution
(a)With mass ${m_2}$ alone, the extension of the spring l is given as ${m_2}g = kl$ …(i)
With mass $({m_1} + {m_2})$, the extension $l'$ is given by
$({m_1} + {m_2})g = k(l + \Delta l)$ ….(ii)
The increase in extension is $\Delta l$ which is the amplitude of vibration. Subtracting (i) from (ii), we get
${m_1}g = k\Delta l$ or $\Delta l = \frac{{{m_1}g}}{k}$
Standard 11
Physics
Similar Questions
normal
