a body of mass o.1 kg is attached to a spring and placed on a horizontal frictionless table. the spring is stretched 20cm,when a force of 5N is applied .what will be the period of oscillation when the mass is set in motion?

25 nm-1

25n m

To find the period of oscillation for a mass-spring system, you need to use Hooke's Law and the equation for the period of simple harmonic motion.

1. Start by applying Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. Mathematically, this can be expressed as:

F = -kx

Where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

2. Rearrange the equation to solve for the spring constant:

k = -F / x

Plugging in the given values, with force F = 5 N and displacement x = 0.2 m (20 cm converted to meters), we get:

k = -5 N / 0.2 m
= -25 N/m

Since the negative sign indicates that the force is acting in the opposite direction of the displacement, we do not need to include it when calculating the period.

3. The equation for the period of oscillation (T) of a mass-spring system is given by:

T = 2π√(m / k)

Where m is the mass of the object attached to the spring, and k is the spring constant.

4. In this case, the mass of the body is 0.1 kg, and the spring constant (k) is -25 N/m. Plugging these values into the equation, we get:

T = 2π√(0.1 kg / -25 N/m)
= 2π√(-0.004 kg/N)

The negative sign indicates that the system is in the attractive (restoring) force region of the spring.

Therefore, the period of the oscillation is given by the equation T = 2π√(-0.004 kg/N).