2. A mass m that is oscillating on a spring with a force constant of 0.78N/m has a period of 3.45 On a second spring the same mass has a period of 5.25. What is the force constant kof the second spring? 3. A 0.15 kg mass stretches a massless spring 0.32 m from its equillbrium position. If this same mass is set into vibration on this spring.then what is the frequency of escillation?

2. The period of oscillation for a mass-spring system is given by the formula T = 2π√(m/k), where T is the period, m is the mass, and k is the force constant. We can rearrange this formula to solve for k: k = 4π²m/T².<br /><br />For the first spring, we have T₁ = 3.45 s and k₁ = 0.78 N/m. Plugging these values into the formula, we get:<br /><br />k₁ = 4π²m/T₁²<br />k₁ = 4π²(0.15 kg)/(3.45 s)²<br />k₁ ≈ 0.78 N/m<br /><br />For the second spring, we have T₂ = 5.25 s and we need to find k₂. Plugging these values into the formula, we get:<br /><br />k₂ = 4π²m/T₂²<br />k₂ = 4π²(0.15 kg)/(5.25 s)²<br />k₂ ≈ 0.32 N/m<br /><br />Therefore, the force constant k of the second spring is approximately 0.32 N/m.<br /><br />3. The frequency of oscillation for a mass-spring system is given by the formula f = 1/2π√(k/m), where f is the frequency, k is the force constant, and m is the mass.<br /><br />For the given massless spring, we have m = 0.15 kg and the spring is stretched 0.32 m from its equilibrium position. We can use Hooke's Law to find the force constant k: F = kx, where F is the force and x is the displacement. Since the mass is in equilibrium, the force is equal to the weight of the mass: F = mg. Therefore, we have:<br /><br />k = F/x<br />k = mg/x<br />k = (0.15 kg)(9.8 m/s²)/(0.32 m)<br />k ≈ 4.69 N/m<br /><br />Now we can plug this value of k into the formula for frequency:<br /><br />f = 1/2π√(k/m)<br />f = 1/2π√(4.69 N/m / 0.15 kg)<br />f ≈ 1.59 Hz<br /><br />Therefore, the frequency of oscillation for the mass on the spring is approximately 1.59 Hz.