A block of mass 200 grams is connected to a light, horizontal spring with a spring constant k= 5 N/m. It is free to oscillate on a frictionless surface, and we’ll also neglect air resistance so you don’t have to worry about decaying. If the block is displaced 5 cm away from equilibrium and released from rest, compute the period Tof its motion.

T = 1.26s

Explanation: From the question,

Mass of loaded spring = 200g = 0.2

Spring constant of spring = 5 N/m

Amplitude = 5cm = 0.05m

The angular frequency (ω) of the harmonic motion of a loaded spring is related to spring constant (k) and mass of loaded spring (m), this is given below as

ω = √k/m

ω = √5/0.2

ω = √25

ω = 5 rad/s

Recall that for a simple pendulum, ω = 2πf where f = frequency of oscillations.

Hence f = ω/2π.

But T = 1/f where T = period

Hence T = 2π/ω

T = 2 × 3.142/ 5

T = 1.26s

T = 1.26 seconds

The period T of its motion is 1.26 seconds

Explanation:

Given:

Mass of block m = 200grams = 0.2kg

Spring constant k = 5N/m

The period P of motion of the spring with mass, can be written as;

T = 2π√(m/k) 1

Where the values of m and k are stated above.

Substituting the values of m and k, we have;

T = 2π√(0.2/5)

T = 2π(0.2)

T = 1.26 seconds

The period T of its motion is 1.26 seconds

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be