Two blocks of masees m 1 and m 2 m 1- m 2 are performing SHM together with same amplitude and same time period as shown- Surface between m - and ground is smooth- while between m 1 and m 2 the coefficient of friction is - Given that k 1-m1- k 2- m 2- choose the correct options A- Time period of SHM is 2 - p - k - where P - m 1 m 2-1-1- m 2 and 1- k -1-k1-1-k2B- Maximum possible amplitude of this SHM is 2 - m 2 g m 1- m 2- k 1 m 2- km 1C- Time period of SHM is 2 - p - k - where P - m 1- m 2 and k - k 1- k 2D- Maximum possible amplitude of this SHM is - m 2 g m 1- m 2- k 1 m 2- k 2 m 1

The correct options are C Time period of SHM is 2 π√(P/k), where P=m_1+m_2 and k=k_1+k_2 D Maximum possible amplitude of this SHM is μm_2g(m_1+m_2)/k_1m_2 k_2m_ ...

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Byju's Answer

Standard XII

Physics

Time Period of Pendulum

Two blocks of...

Question

Two blocks of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are performing SHM together with same amplitude and same time period as shown. Surface between $m_{1}$ and ground is smooth, while between $m_{1}$ and $m_{2}$ the coefficient of friction is $μ$ . Given that $\frac{k_{1}}{m_{1}} > \frac{k_{2}}{m_{2}},$ choose the correct option(s)

Time period of SHM is

2 π \sqrt{\frac{P}{k}},

where

P = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

and

\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}}

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Maximum possible amplitude of this

S H M

\frac{2 μ m_{2} g (m_{1} + m_{2})}{k_{1} m_{2} - k_{2} m_{1}}

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Time period of SHM is

2 π \sqrt{\frac{P}{k}},

where

P = m_{1} + m_{2}

and

k = k_{1} + k_{2}

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Maximum possible amplitude of this

S H M

\frac{μ m_{2} g (m_{1} + m_{2})}{k_{1} m_{2} - k_{2} m_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
C Time period of SHM is $2 π \sqrt{\frac{P}{k}},$ where $P = m_{1} + m_{2}$ and $k = k_{1} + k_{2}$
D Maximum possible amplitude of this $S H M$ is $\frac{μ m_{2} g (m_{1} + m_{2})}{k_{1} m_{2} - k_{2} m_{1}}$
Consider $m_{1}$ and $m_{2}$ as whole system, since the springs are parllel $k = k_{1} + k_{2} .$

Let $a_{m a x}$ be the maximum acceleration of system
$⟹ (m_{1} + m_{2}) \times a_{m a x} = (k_{1} + k_{2}) \times A$ (A is amplitude of system)

We know $f_{m a x} = μ m_{2} g$

From figure above
$f_{max} + k_{2} A = m_{2} (\frac{k_{1} + k_{2}}{m_{1} + m_{2}}) A$
$μ m_{2} g = [m_{2} (\frac{k_{1} + k_{2}}{m_{1} + m_{2}}) - k_{2}] A$

$A = \frac{μ m_{2} g (m_{1} + m_{2})}{k_{1} m_{2} - k_{2} m_{1}}$

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Similar questions

Q. Two blocks of masses

m_{1}

and

m_{2}

(

m_{1} > m_{2}

), are performing SHM together with the same amplitude and same time period as shown. Surface between

m_{1}

and the ground is smooth, while between

m_{1}

and

m_{2}

, the coefficient of friction is

μ

. Given that

\frac{k_{1}}{m_{1}} > \frac{k_{2}}{m_{2}}

. Choose the correct option:

Q. In the arrangement shown, both the springs are in their natural lengths. The coefficient of friction between

m_{2}

and

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μ

. There is no friction between

m_{1}

and surface. If the blocks are displaced slightly, they perform SHM together.
then, the
(1) frequency of such oscillations

\frac{1}{2 π} \sqrt{\frac{k_{1} + k_{2}}{m_{1} + m_{2}}},

(2) condition if the frictional force on block

m_{2}

is to act in the direction of its displacement from the mean position is

\frac{m_{1}}{m_{2}} > \frac{k_{1}}{k_{2}}

(3) If the condition obtained in (b) is met, the maximum amplitude of their oscillations is

\frac{μ m_{2} g (m_{1} + m_{2})}{k_{2} m_{1} - k_{1} m_{2}}

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