A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use energy ?conservation to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

Solution 37E Step 1: Introduction : In this question ,we need to find the amplitude of the motion In the second part we need to find the maximum speed of the glider In the third part we need to find the angular frequency of the oscillations Data given Mass of the glider m = 175 g = 0.175 kg Force constant k = 155 N/m Velocity v = 0.815 m/s Distance x = 3.0 cm Step 2 : When the glider is at rest, all the energy stored in the glider accomplies to potential energy PE Whereas when the glider is moving with maximum velocity, the energy of the glider is given by Kinetic energy Hence using law of conservation of energy we have KE = PE We can write this as KE = 1/2 mv and PE = 1/2 kx 2 Since the total energy E of the system is given by E = 1/2 mv + 1/2 kx 2 Substituting values we get E = 1/2 × 0.175 kg × (0.815 m/s) + 1/2 × 155 N/m × (0.03 m) 2 E = 0.0581 J + 0.06975J E = 0.12785 J This can be approximated to E = 0.128 J Hence we have the total energy of the glider as 0.128 J Step 3 : We can obtain the amplitude using A = 2E/K Substituting values we get A = 2 × 0.128 J)/ 155 N/m A = .65 × 10 3 A = 0.0406 m This can be written as A = 4.06 cm Hence we have the amplitude as 4.06 cm