Question

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 0.120 m. The mass is now displaced from its equilibrium position and undergoes simple harmonic oscillations. How long does it take the mass to complete one full oscillation?

Answer 1

The time it takes for a 0.350 kg mass attached to a spring to complete one full oscillation is found to be approximately 1.17 seconds, using the period formula for simple harmonic motion and calculating the spring constant using Hooke's Law.

Step-by-step explanation:

The question asks about the period of simple harmonic motion for a mass attached to a spring. The period of oscillation (T) for a mass-spring system can be determined using the formula T = 2π√(m/k), where 'm' is the mass of the object and 'k' is the spring constant. Since we are given the mass (0.350 kg) and the amount the spring stretches due to this mass (0.120 m), we can use Hooke's Law, F = -kx, to find the spring constant 'k' as F/mg, where 'F' is the force exerted by the mass (which is its weight) and 'x' is the displacement (spring stretch). Calcualting this we find:

k = mg/x = (0.350 kg × 9.8 m/s²) / 0.120 m = 28.75 N/m

Now, by plugging 'm' and 'k' into the formula for the period, we get:

T = 2π√(0.350 kg / 28.75 N/m) ≈ 1.17 s

Therefore, it takes approximately 1.17 seconds for the mass to complete one full oscillation.

Answer 2

Time period of the oscillation will be 0.695 sec

Step-by-step explanation:

We have given mass attached to the string m = 0.350 kg

The spring is stretched by 0.120 m

So x = 0.120 m

We know that force is given by F = Kx

So mg = kx

`0.350* 9.8=k* 0.120`

k = 28.5833

Now time period is given by

`T=2\pi \sqrt{(m)/(k)}=2* 3.14\sqrt{(0.35)/(28.5833)}=0.695sec`

So time period of the oscillation will be 0.695 sec