Question

In the diagram shown blocks are oscillating harmonically. There is no friction between blocks and floor. The coefficient of friction between m2 and m3 is ‘u’. The velocity of centre of mass is zero. The maximum amplitude of oscillation of m1 for which m2 does not slip on m3 is

Answer 1

Answer:

Amplitude of mass m1 is given as

$A_1 = \frac{(m_2 + m_3)^2 \mu g}{k(m_1 + m_2 + m_3)}$

Explanation:

For sliding of m2 block on the surface of m3 we know that the friction force between m2 and m3 will reach to its limiting friction value

So we have

$F_f = \mu m_2 g$

Since we know that the acceleration of the block is maximum at its maximum compression or maximum extension position

So we have

$m_2a_2 = \mu(m_2 g)$

$a_2 = \mu g$

now we know that the system will have angular frequency given by the concept of reduced mass

so we have

Now we know that COM is at rest and both parts will oscillate about it with amplitude A1 and A2 so we have

$k_1A_1 = k_2A_2$

now we know that

$k_2A_2 = (m_2 + m_3) a_{max}$

$k_2 A_2 = (m_2 + m_3)\mu g$

so we have

$k_1A_1 = k_2A_2 = (m_2 + m_3)\mu g$

now we know that

$k_1 = \frac{(m_1 + m_2 + m_3)}{m_2 + m_3} k$

so we have

$A_1 = \frac{(m_2 + m_3)^2 \mu g}{k(m_1 + m_2 + m_3)}$

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Topic : SHM

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