4.4 Solve the equation of motion for the inverted simple pendulum from the previous problem for small deviations from vertical. How does the solution differ from that for the simple pendulum (i. e., simple harmonic motion)?

Light from a sodium lamp passes through a diffraction grating that has 1000 slits per millimeter. the interference pattern is viewed on a screen 1.000 m behind the grating. the first (m = 1) two bright yellow fringes that are visible are 0.7288 m and 0.7300 m from the central maximum. what are the wavelengths of these two fringes?

A0.580-kg rock is tied to the end of a string and is swung in a circle with a radius of 0.500 meters. the velocity of the rock is 4.50 m/s. what is the centripetal force acting on the rock? 15.5 n 5.22 n 69.8 n 23.5 n

Calculate the first and second velocities of the car with four washers attached to the pulley, using the formulas v1 = 0.25 m / t1 , and v2 = 0.25 m / (t2 – t1) where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. record these velocities, to two decimal places, in table e.