1. A 0.600 kg mass is hung from a 25N/m spring. What is its period of oscillation? 2. A 10 kg mass is hung from a spring. It oscillates up and down 15 times in 10 seconds. a. What is its period of oscillation? b. What is the spring constant? 3. A pendulum is 0.5 m long. What is its period of oscillation? 4. A pendulum swings back and forth 5 times in 10 seconds. a. What is its period of oscillation? b. How long is the pendulum? n:11 in the following table milk it

Let's solve each problem step by step:<br /><br />1. **Mass-Spring System: Period of Oscillation**<br /><br /> The period \( T \) of a mass-spring system is given by the formula:<br /> \[<br /> T = 2\pi \sqrt{\frac{m}{k}}<br /> \]<br /> where \( m \) is the mass and \( k \) is the spring constant.<br /><br /> Given:<br /> - Mass \( m = 0.600 \, \text{kg} \)<br /> - Spring constant \( k = 25 \, \text{N/m} \)<br /><br /> Plug these values into the formula:<br /> \[<br /> T = 2\pi \sqrt{\frac{0.600}{25}} = 2\pi \sqrt{0.024} \approx 2\pi \times 0.155 \approx 0.974 \, \text{seconds}<br /> \]<br /><br />2. **Mass-Spring System: Frequency and Spring Constant**<br /><br /> a. **Period of Oscillation**<br /><br /> The frequency \( f \) is given as 15 oscillations in 10 seconds, so:<br /> \[<br /> f = \frac{15}{10} = 1.5 \, \text{Hz}<br /> \]<br /><br /> The period \( T \) is the reciprocal of the frequency:<br /> \[<br /> T = \frac{1}{f} = \frac{1}{1.5} \approx 0.667 \, \text{seconds}<br /> \]<br /><br /> b. **Spring Constant**<br /><br /> Using the formula for the period of a mass-spring system:<br /> \[<br /> T = 2\pi \sqrt{\frac{m}{k}}<br /> \]<br /><br /> Rearrange to solve for \( k \):<br /> \[<br /> k = \frac{4\pi^2 m}{T^2}<br /> \]<br /><br /> Given:<br /> - Mass \( m = 10 \, \text{kg} \)<br /> - Period \( T = 0.667 \, \text{seconds} \)<br /><br /> Plug these values into the formula:<br /> \[<br /> k = \frac{4\pi^2 \times 10}{(0.667)^2} \approx \frac{39.478 \times 10}{0.444} \approx 890.18 \, \text{N/m}<br /> \]<br /><br />3. **Pendulum: Period of Oscillation**<br /><br /> The period \( T \) of a simple pendulum is given by:<br /> \[<br /> T = 2\pi \sqrt{\frac{L}{g}}<br /> \]<br /> where \( L \) is the length of the pendulum and \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.<br /><br /> Given:<br /> - Length \( L = 0.5 \, \text{m} \)<br /><br /> Plug this value into the formula:<br /> \[<br /> T = 2\pi \sqrt{\frac{0.5}{9.81}} \approx 2\pi \times 0.225 \approx 1.419 \, \text{seconds}<br /> \]<br /><br />4. **Pendulum: Frequency and Length**<br /><br /> a. **Period of Oscillation**<br /><br /> The frequency \( f \) is given as 5 oscillations in 10 seconds, so:<br /> \[<br /> f = \frac{5}{10} = 0.5 \, \text{Hz}<br /> \]<br /><br /> The period \( T \) is the reciprocal of the frequency:<br /> \[<br /> T = \frac{1}{f} = \frac{1}{0.5} = 2 \, \text{seconds}<br /> \]<br /><br /> b. **Length of the Pendulum**<br /><br /> Using the formula for the period of a pendulum:<br /> \[<br /> T = 2\pi \sqrt{\frac{L}{g}}<br /> \]<br /><br /> Rearrange to solve for \( L \):<br /> \[<br /> L = \left(\frac{T}{2\pi}\right)^2 \times g<br /> \]<br /><br /> Given:<br /> - Period \( T = 2 \, \text{seconds} \)<br /><br /> Plug this value into the formula:<br /> \[<br /> L = \left(\frac{2}{2\pi}\right)^2 \times 9.81 \approx (0.318)^2 \times 9.81 \approx 0.1008 \times 9.81 \approx 0.989 \, \text{m}<br /> \]<br /><br />These are the solutions to the problems provided.