Question

Please help me understand this. I know answers are already out there but please don't simply copy and paste. i'm looking for explanations a

23. Consider the following transverse wave and Y(z,t) = 20sm(TI-π). Please calculate the function of the transverse speed. Must show work for credit. onsider the following transverse velocity wave and V(x,t the function of the transverse acceleration. Must show work for credit. 25. A wave propagates in a string. The tension on the sting is 100N while the linear density μ is 1Kg/m. Please calculate the velocity of the propagation. Must show work for credit.

Answer 1

Given that (23) Y(x, t) = 20 sin(pi x - pi t) nu(x . t) = d[y(x, t)]/dt = d/dt[20 sin(pi x - pi t)] = 20 cos(pi x - pi t) (-pi(1) Therefore nu(x, t) = -20pi cos(pi x - pi t) Therefore d/dt [sin(a + bt)] = cos(a + bt)(a + b) = bcos(a + mbt) (24) v(x, t) = cos(2pix - 3pi t) a(x, t) = dV(x, t)/dt = d/dt [cos(2pi x - 3pi t)] = -sin(2pi x - 3 pi t) d/dt (2 pu x - 3 pi t) = -sin (2pi x - 3 pi t)(0 - 3 pi (1)] Therefore a(x, t) = 3 pi sin(2 pi x - 3 pi t) (25) T = 100 N , mu = 1 kg/m The velocity of propagation (nu) = squareroot T/u = squareroot 100/1 = 10 material implication nu = 10 m/s Therefore nu = 10. m/s.