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P a Spring Stores 5 J of Energy When Stretched by 25 Cm. It is Kept Vertical with the Lower End Fixed.

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प्रश्न

A spring stores 5 J of energy when stretched by 25 cm. It is kept vertical with the lower end fixed. A block fastened to its other end is made to undergo small oscillations. If the block makes 5 oscillations each second what is the mass of the block?

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उत्तर

It is given that:
Energy stored in the spring, E = 5 J
Frequency of the mass-spring system, = 5
Extension in the length of the spring, = 25 cm = 0.25 m

\[\text { Time  period },   T   = \frac{1}{5}  s\] 

\[\text { Potential  energy }\left( U \right)\text {  is  given  by, }  \] 

\[U = \frac{1}{2}k x^2 \] 

\[ \Rightarrow \frac{1}{2}k x^2  = 5\] 

\[ \Rightarrow \frac{1}{2}k \left( 0 . 25 \right)^2  = 5\] 

\[ \Rightarrow k = 160  N/m\] 

\[\text { Time  period  of  spring  mass  system  is  given  by, }\] 

\[  T = 2\pi\sqrt{\left( \frac{m}{k} \right)}\] \[ \text {where  m  is  the  mass  of  the  body  hanged,   and }\] \[\text { k  is  the  spring  constant . }\]\[\text { On  substituting  the  respective  values  in  the  above  expression,   we  get: }\]

\[      \frac{1}{5} = 2\pi\sqrt{\left( \frac{m}{160} \right)}\] 

\[ \Rightarrow m = 0 . 16  kg\]

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पाठ 12: Simple Harmonics Motion - Exercise [पृष्ठ २५३]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
पाठ 12 Simple Harmonics Motion
Exercise | Q 13 | पृष्ठ २५३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

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