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The Block of Mass M1 Shown in Figure (12−E2) is Fastened to the Spring and the Block of Mass M2 is Placed Against It. - Physics

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प्रश्न

The block of mass m1 shown in figure is fastened to the spring and the block of mass m2 is placed against it. (a) Find the compression of the spring in the equilibrium position. (b) The blocks are pushed a further distance (2/k) (m1 + m2)g sin θ against the spring and released. Find the position where the two blocks separate. (c) What is the common speed of blocks at the time of separation?

बेरीज
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उत्तर

(a) As it can be seen from the figure,
Restoring force = kx
Component of total weight of the two bodies acting vertically downwards = (m1 + m2sin θ
At equilibrium,  
 kx = (m1 + m2sin θ

\[\Rightarrow x = \frac{\left( m_1 + m_2 \right)  g  \sin  \theta}{k}\]

(b) It is given that:
Distance at which the spring is pushed,

\[x_1  = \frac{2}{k}\left( m_1 + m_2 \right)g  \sin  \theta\]

As the system is released, it executes S.H.M.
where \[\omega = \sqrt{\frac{k}{m_1 + m_2}}\]

When the blocks lose contact, becomes zero.                   (is the force exerted by mass m1 on mass m2)

\[\therefore    m_2 g  \sin  \theta =  m_2  x_2  \omega^2  =  m_2  x_2  \times \frac{k}{m_1 + m_2}\] \[ \Rightarrow  x_2  = \frac{\left( m_1 + m_2 \right)  g  \sin  \theta}{k}\]

Therefore, the blocks lose contact with each other when the spring attains its natural length.

(c) Let v be the common speed attained by both the blocks.

\[\text { Total  compression }  =    x_1  +  x_2 \] 

\[\frac{1}{2}  \left( m_1 + m_2 \right) v^2  - 0   =   \frac{1}{2}k \left( x_1 + x_2 \right)^2  - \left( m_1 + m_2 \right)  g  \sin  \theta  \left( x + x_1 \right)                      \] 

\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2  = \frac{1}{2}k\left( \frac{3}{k} \right)  \left( m_1 + m_2 \right)  g  \sin  \theta - \left( m_1 + m_2 \right)  g  \sin  \theta  \left( x_1 + x_2 \right)\] 

\[ \Rightarrow \frac{1}{2}\left( m_1 + m_2 \right) v^2  = \frac{1}{2}  \left( m_1 + m_2 \right)  g  \sin  \theta \times \left( \frac{3}{k} \right)  \left( m_1 + m_2 \right)  g  sin  \theta\] 

\[ \Rightarrow v = \sqrt{\left\{ \frac{3}{k}  \left( m_1 + m_2 \right) \right\}}g  \sin  \theta\]

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Energy in Simple Harmonic Motion
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५३]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
पाठ 12 Simple Harmonics Motion
Exercise | Q 15 | पृष्ठ २५३

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