Question

A horizontally
kept linear elastic spring having a spring constant 'K'is fixed at one end and has
an object of mass 'm' attached to its free end. The mass 'm' is stretched from the equilibrium
position through a distance x = a. Obtain an expression for the potential energy of the spring in
temps of its displacement X (O< X <a). Also find the total energy of the spring at any position
Obtain expression for the speed of the mass 'm', at the displacement 'x', when
(1) 0<x<a
(2) x = 0
​

Answer 1

Answer:

Part a) 0 < x < a

Potential energy = $\frac{1}{2}kx^2$

Total energy = $\frac{1}{2}ka^2$

Part b) x = 0

Potential energy = 0

Total energy = $\frac{1}{2}ka^2$

Explanation:

As we know that potential energy is defined as

$U = \frac{1}{2}kx^2$

where x = displacement from mean position

so we know that

i) 0< x < a

Potential energy

$U = \frac{1}{2}kx^2$

Total energy is always constant

$E = \frac{1}{2}ka^2$

ii) x = 0

when x = 0 so spring is at its mean position

so we have

$U = 0$

Total energy is always constant

$E = \frac{1}{2}ka^2$

#Learn

Topic : Energy stored in the spring

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