Let there be a spring as shown. And this is stretched to point C and released, then force exerted by spring on mass in region X2 is "- kx" . But This springs gains kinetic energy at mean position and moves till P. Could you please tell me the force expression (force produced by spring on mass) in X1 region i.e., at pt. A at some distance x. Assume suitable variables and massless mass attached to spring. 
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2 Answers
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I think you meant to say the spring is massless, correct me if I'm wrong.
Force on the mass by string if we assume down direction as positive ( to follow your sign -kx pulling up) is $kx, \ $ it is pushing the mass down.
At the same time, the force exerted by the mass on the spring is $-kx$.
The mg component is permanently pulling the mass down and as far as inertia it doesn't come into play. All it does is moves the equilibrium position down by an amount $$x_{equilibrium}= \frac{mg}{k}$$
And the mass is experiencing two forces $mg+kx$
Usually, we set the reference point at the end of the string where it attaches to the mass by the hook, not at the middle, but it doesn't make a difference.
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$\begingroup$ Thank you for your response! Let me make my question more clear. The spring force in region X1 is downwards, then why does the spring move upwards. I am assuming it's due to the inertial effect of spring. So I am looking for the expression for force in X1 which is causing the spring to go upwards (besides downward force by spring). $\endgroup$– KaneCommented Jan 27, 2022 at 18:34
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1$\begingroup$ the spring goes upward because of the inertia of the mass =1/2 mv^2. in a spring-mass harmonic vibration, the mass is trading kinetic energy to potential energy and back in each cycle. it has zero kinetic energy on top and bottom but has loaded the spring. in equilibrium point the mass is at max speed, max kinetic energy but the spring is relaxed, it has given all it energy and is unstretched. in our model, we ignore any spring's inertia. otherwise, we had to deal with continuum mechanics and tensors. $\endgroup$– kamranCommented Jan 27, 2022 at 18:46
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Assuming that:
- x=0 is the point where the spring is neither compressed nor extended
- x is positive upwards
- gravity plays a role then the motion equation is:
$$m\cdot \ddot{x} + k\cdot x = -m\cdot g$$
By redistributing the terms:
$$m\cdot \ddot{x} = -m\cdot g - k\cdot x $$
Because the first term ($m\cdot \ddot{x}$) is usually considered inertial force then what you are after is:
$$ \text{inertial force}= -m\cdot g - k\cdot x $$
-kx(x, of course has a different sign in region X1 compared to region X2) ? Why do you think that the answer is something else? $\endgroup$inertialin it? Are you not looking for the spring force? $\endgroup$