Question

A string vibrates according to the equation 1. Y(x,t) = 10.0 cm * sin((7.00cm-')x)*cos(2.00 rad/s)t (THIS IS A STANDING WAVE. What is the angular wave number? What is the wavelength? What is the speed of the 2 waves making the standing wave? What is the amplitude of the 2 waves making the standing wave? What is the distance between the nodes? What is the angular speed of the 2 waves making the standing wave? What is the period of the 2 waves making the standing wave? What is the maximum transverse speed of this standing wave? Write the equations for the two waves making up the standing wave Y1(x,t) = Y2(x,t) = 2. A disk of mass m = 8.00 kg and radius r rolls without slipping along a loop-the-loop track, with a height of h and radius R=10 m. The disk is released from rest somewhere on the straight section of the track. a. What is the beta value of a disk? b. From what minimum height h above the bottom of the track must the disk be released in order that it not leave the track at the top of the loop? c. What is the potential energy at the start? h 2R

Answer 1

y havenumber tomt Ano equation of standing wave! yox,t) = 10.om Sm Cez.com?)a) (GC200 XW10) 5) Compare it with 2 A Sonk a Cesat. A = 5cm k=7cml w=dico radio k= 277 Fix=271 cm. And D Wavelength = 27 cm Ano ware speed = v= co= a emis Ano Amplitude A = Som Ano distance between two modes 2 = 77 cm. Ara 7

- Tranolati malkmeticenery MR2 B= 2 Iw2 mue moz Anguler speed w=aradio. Ass Time period T= 20 = T 8 Te Amo Maxinum transverse speeds AW =5x2 =10cm. Moment of Inestra of disc I= MR2 2 I= 8x(1032 Lookg ma liciget h. Bralece= Rotation kmentic energyon elloy 놀 b. The Look should not leave the track at the I mus=Yq = ß=1/2 Ano lightest bont le wtsheuld be balance by balanced bycentrifugel force at tok yar yg =) UzdRg. energy Noto total here ) 22w²+Lmoe = mos (1+8 = 3 mua Now by energy conservation total energy at bottom or mital total energy. £ mus ² (1+B) friscal at ligbest pornit TEJ mor (1+B) + mgaR = 3 mus+mgar Now By energy conservation! TE=TEJ 8 must = moet ngar R Img 1 2 Total energy V2 JYR

= of 3 A Umumi 2002= 3h gR+992R 3002 119R i Vo=dligr And 19R 3 Now by energy conser ration highest pontot motion and bottom: mga=3 mus²=110gr 4 h= 112 = 11810 11R 2.7.5m 4 A - hmoni (onininum beeight) = 27.5m Ano Now potential energy at this ponit- PE = mgh=8x98x27.5 =2156 J And Remang part of lot quent Yo Cam B) A Borckx-cot) = 58on ( 72-26) Ang Y 2 @ 1 t) - Asom Clexet cot) =560 & #425) Ano

sorry for the bad hand writting if you can not understand any thing please comment i will explain it furthur thank you.