Question

This scenario is for questions 1-2 A simple harmonic oscillator at the position x-generates a wave on a string. The oscillator moves up and down at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. At time t=0, the oscillator is passing through the origin and moving down. The string has a linear mass density of 50,0 g/m and Is stretched with a tension of 5.00 N. a) Find the angular frequency of the oscillator. b) If we express the oscillator's motion as y(t) = A cos(wt + f), what is the phase anglef? c) At what time t will the oscillator be at its maximum positive displacement of 3.00 cm? d) Find the speed of the wave down the string. e) Find the wavelength of the wave on the string. f) Draw a snapshot of the wave att-2.5 s. Please put numbers on both axes.

could you help me with g-j please?

Answer 1

f= 40 HZ Amplitude A= 3x10²m mass density = soxio3 kg/m Tension I-5N (a) angular frequency و علما = 2X3.14x40 w =251-2 roots (b) y = A cos (W++6) the Oscillata passing through origin at to 0= Aces (0+¢) Cos&so as a &=512 so phase angle is 1/2

ce) y-Acos (W7+6) W-251.2 radys but Amplitude A=3 cm y = 3cm maxianu displacement 3= 3 Col (951.2t + H 2 ) Cos (25112t+72) = = Ces27 251 2t + 5/2 = 21 251.2t= 25-2/2 = 32 t = 3X3.14 0.019 second. 2X251.2 (d) Velocity / speed of wave down the string where t is tension in string and mis massper unit length m=so so x103 kg/m 12 m T= 54 v= 500 Toms. ==x v=fd + = x - to = 0.25m

(o y - A. Cal Cutte) y=365 (25112t+42) at t-2.55 y = 368 257.2 X 2.5+77) y = 360$ (628+72) y = - 3 Sin 628 =-3X (-0.94 y = 2 - 2.97 cm ch) ody=-Awsin (wtto] dt velocity v= dy at t = 25 second V=-3x251:2 Sin 251.2X2:57 v=-753-6 Sin (62272/2) V = -753.6 cs 628 =-753.6% +0.025] TE 26.38 chys = 0.26 m als (۱) So a= dy - dv = acceleration dt Awe costutte) a=-why - (25112) ² (-2 sin628) (251.2.3" x 2.97 = 1.87 XI0 a = 1:87X10"mis? saus? a=