i need the answer quickly The wave function for a standing wave on astring of linear mass density u = 0.2 kg/m, isgiven by y(x,t) = 0.05 sin(4Ttx)cos(60rtt),where x and y are in meters and t is inseconds. At what time would an element ofthe string, located at x 0.1 m, has itstransverse velocity v_y = v_(y,max)/2?O t= T/60 secO t= n/360 secO t= 1/360 sect = 1/60 sec

since at any point V=dydtVy=-Vymaxsin60πt

The wave function for a standing wave on a string of linear mass density u = 0.2 kg/m, is given by y(x,t) = 0.05 sin(4rtx)cos(60rt), where x and y are in meters and t is in seconds. At what time would an element of the string, located at x 0.1 m, has its transverse velocity v_y = v_(y,max)/2? O t= T/60 sec O t= n/360 sec O t= 1/360 sec t = 1/60 sec

The wave function for a standing wave on a string of linear mass density u = 0.2 kg/m, is given by y(x,t) = 0.05 sin(4rtx)cos(60rt), where x and y are in meters and t is in seconds. At what time would an element of the string, located at x 0.1 m, has its transverse velocity v_y = v_(y,max)/2? O t= T/60 sec O t= n/360 sec O t= 1/360 sec t = 1/60 sec

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter17: Sound

Section: Chapter Questions

Problem 68P: The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor...

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