One end of a horizontal string of linear density 4.21E-4 kg/m is attached to a s

Question

One end of a horizontal string of linear density 4.21E-4 kg/m is attached to a small-amplitude mechanical 53.3 Hz vibrator. The string passes over a pulley, a distance L = 1.30 m away, and weights are hung from this end, as seen in the figure below.Assume the string at the vibrator is a node, which is nearly true.

Q:What mass must be hung from this end of the string to produce 1 loop of a standing wave?

Q:What mass must be hung from this end of the string to produce 2 loops of a standing wave?

Q:What mass must be hung from this end of the string to produce five loops of a standing wave?

One end of a horizontal string of linear density 4.21E-4 kg/m is attached to a small-amplitude mechanical 53.3 Hz vibrator. The string passes over a pulley, a distance L = 1.30 m away, and weights are hung from this end, as seen in the figure below.Assume the string at the vibrator is a node, which is nearly true. Q:What mass must be hung from this end of the string to produce 1 loop of a standing wave? Q:What mass must be hung from this end of the string to produce 2 loops of a standing wave? Q:What mass must be hung from this end of the string to produce five loops of a standing wave?

Explanation / Answer

Basic equation: f = sqrt(T/(m/L))/(2L)

==> T = (2L)^2*(m/L)
A. T(1) = 11.85408 N , now M = T(1)/g

so M = 1.208kg
B For 2 loops f(2) = 2*56 Hz , Frequency is proportional to T^2, thus T(2) = 2^2*T(1)

thus T(2) = 4*11.85408 N ... dividing it by g we get

M = 4.83 kg

C. For 5 loops, f(5) = 5*56 Hz. Frequency is proportional to T^2, thus T(5) = 5^2*T(1).
T(5) =25*11.85408 N ... again dividing by g we get

M = 30.309 kg

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