The condition, if the frictional force on block $m_{2}$ is to act in the direction of its displacement from the mean position is met, what can be maximum amplitude of their oscillations?

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Solution

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

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Similar questions

Q. The condition if the frictional force on block

m_{2}

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

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

m_{2}

and

m_{1}

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}}

Q. A particle is describing SHM with amplitude

^{'} a^{'} .

When the potential energy of particle is one fourth of the maximum energy during oscillation, then its displacement from mean position will be:

A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude ? (c) What can be the maximum amplitude with which the two blocks may oscillate together ?

Q. The maximum displacement of the particle from its mean position is called amplitude.

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The condition, if the frictional force on block m2 is to act in the direction of its displacement from the mean position is met, what can be maximum amplitude of their oscillations?

k1Am+μm2g=m1(k1+k2m1+m2)AmAm(m1k1+m1k2m1+m2−k1)=μm2gor Am=μ(m1+m2)m2gm1k2−m2k1

The condition, if the frictional force on block $m_{2}$ is to act in the direction of its displacement from the mean position is met, what can be maximum amplitude of their oscillations?

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