EPHY111L G20 Lecture 14 Mechanics (Bennett University) PDF

MECHANICS EPHY111L G20 Lecture 14 Arindam Lala Dept. of Physics Bennett University  Coupled Oscillator  Let two mass points are attached to walls by two 𝑚1 = 𝑀 𝑚2 = 𝑀...

MECHANICS EPHY111L G20 Lecture 14 Arindam Lala Dept. of Physics Bennett University  Coupled Oscillator  Let two mass points are attached to walls by two 𝑚1 = 𝑀 𝑚2 = 𝑀 springs with same spring constants and  The two similar masses and are coupled by 𝑘 1= 𝑘 𝑘12 𝑘 2= 𝑘 another spring of spring constant 𝑥1 𝑥2  We take the motion along the horizontal -axis only and are displacements about the The equations of motion of the two masses equilibrium positions of the two in masses a rr ang Re g  We want oscillatory solutions Substitut  Assume e Trial solution s  Mathematical Analysis A pair of simultaneou s equations For a solution to exist, the determinant of the coefficients of and must vanish Hence, Solving for : Two characteristic frequencies of the system  A change in coordinate trick !! Let us define two new coordinates: and Solving: and Substituting into We get Still coupled equations We add and Two uncoupled subtract independent √ equations General solutions are 𝜔 1= 𝑘 +2 𝑘12 𝑀 𝐚𝐧𝐝 𝜔 2= and are called normal √ coordinates  Anti-symmetrical mode Symmetrical mode (out of phase) (in phase) for all for all The particles oscillates always The particles oscillates in-phase out-of-phase with frequency with frequency  If we were to hold fixed and allow to oscillate, the oscillation frequency would have been  Due to symmetry, this would have been the same frequency of oscillation if we held fix and allowed to oscillate  Let be the frequency for uncoupled motion – then  For number of coupled oscillators, characteristic frequencies are always greater than and characteristic frequencies are always less than  For number of coupled oscillators, characteristic frequencies are always greater than and ( characteristic frequencies are always less than and one frequency is equal to 𝜔1 𝜔1 𝜔0 𝜔0 𝜔0 𝜔2 𝜔2 𝑛=2 𝑛=3 Initial velocity = 30 m/sec Radius of the boomerang = Mass of the boomerang = Rotation of the boomerang = Moment of inertia = (a) Total energy of the boomerang when it leaves the hand ; (b) Maximum height attained by the boomerang from the elevation of the hand.