Can you help me with this? I keep getting the wronganswers? One end of a horizon

Question

Can you help me with this? I keep getting the wronganswers?
One end of a horizontal string is attached to a small-amplitudemechanical 62 HZ vibrator. The string's ,ass per unit lengthis 4.0X10-4 kg/m. The string passes over a pulley,a distance L=1.50m away, and weights are hung from one end. Assume the string at the vibrator is a node, which is nearlytrue. What is the mass m that must be hung from this end ofthe string to produce a standing wave with the following number ofloops? 1 loop, 2 loops, and 5 loops.

Explanation / Answer

Frequency f= 62 Hz mass per unit length = 4 * 10^-4 kg / m Length L = 1.5 m For 1 loop : ------------ f = ( 1/2L ) [ Mg / ] 62 = ( 1/3) [M * 24500 ]      = 52.174 M M = 1.412 Kg For 2 loops : ------------- f = ( 2/2L ) [ Mg / ] 62 = ( 2/3) [M * 24500 ]      = 104.348 M M = 0.353 Kg For 5 loops: ------------ f = ( 5/2L ) [ Mg / ] 62 = ( 5/3) [M * 24500 ]      = 260.87 M M = 0.05648 Kg 62 = ( 2/3) [M * 24500 ]      = 104.348 M M = 0.353 Kg For 5 loops: ------------ f = ( 5/2L ) [ Mg / ] 62 = ( 5/3) [M * 24500 ]      = 260.87 M M = 0.05648 Kg 62 = ( 5/3) [M * 24500 ]      = 260.87 M M = 0.05648 Kg

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