A standing wave on a 1-m string fixed at both ends is described by: y(x.t) = 0.3sin(5Ttx)cos(60rtt), where x and y are in meters and t is in seconds. An element on the string located at x = 0.1m would have a maximum transverse speed for the third time at: O t= 1/40 sec O t= 1/60 sec O t= 1/20 sec. O t= 1/24 sec O t= 1/30 sec
A standing wave on a 1-m string fixed at both ends is described by: y(x.t) = 0.3sin(5Ttx)cos(60rtt), where x and y are in meters and t is in seconds. An element on the string located at x = 0.1m would have a maximum transverse speed for the third time at: O t= 1/40 sec O t= 1/60 sec O t= 1/20 sec. O t= 1/24 sec O t= 1/30 sec
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![A standing wave on a 1-m string fixed at both ends is described by: y(x.t) 3D
0.3sin(5Ttx)cos(60rtt), where x and y are in meters and t is in seconds. An
element on the string located at x = 0.1 m would have a maximum transverse
speed for the third time at:
O t-1/40 sec
Ot-1/60 sec
O t- 1/20 sec
O t- 1/24 sec
O t=1/30 sec](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffd3f208d-787d-4277-8cfb-4c72679cea42%2F55dd11ca-97a4-4e83-bcd3-4a4f83853193%2Fsja6bk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A standing wave on a 1-m string fixed at both ends is described by: y(x.t) 3D
0.3sin(5Ttx)cos(60rtt), where x and y are in meters and t is in seconds. An
element on the string located at x = 0.1 m would have a maximum transverse
speed for the third time at:
O t-1/40 sec
Ot-1/60 sec
O t- 1/20 sec
O t- 1/24 sec
O t=1/30 sec
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