How Much Mass to Hang for Different Standing Wave Loops?

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Homework Help Overview

The discussion revolves around a physics problem involving standing waves on a string attached to a mechanical vibrator. The original poster seeks to determine the mass required to achieve different numbers of loops in the standing wave, given specific parameters such as frequency and string mass per unit length.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between wave frequency, tension, and mass. Some attempt to derive equations related to wave speed and tension, while others express confusion about the wave equation and its application to the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance without offering direct solutions. There is a focus on breaking down the problem into manageable steps, and various interpretations of the wave properties are being explored.

Contextual Notes

Participants note the need for prior working to receive assistance, indicating a homework policy that emphasizes independent problem-solving. The original poster's assumptions about the string's behavior at the vibrator are also under consideration.

chazgurl4life
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One end of a horizontal string is attached to a small-amplitude mechanical 57 Hz vibrator. The string's mass per unit length is 4.4 10-4 kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end Assume the string at the vibrator is a node, which is nearly true.

What mass m must be hung from this end of the string to produce a standing wave with the following number of loops?
(a)one loop-----kg?

(b) two loops----------kg?

(c) five loops -------------kg?
 
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Hi chazgurl, you should have realized by now that we do not hand out help or solutions unless you have shown some prior working.

HINTS: (1) Equation for the velocity of a wave on a string.
(2) Wave equation.

~H
 
frequency= 1/2pi(g/L)^1/2= so...-->4.02 Hz

vo=2*pi*frequency*amplitude=

v=+/-Vo ( 1-x^2/A^2)^1/2


i don't know the wave equation
 
chazgurl4life said:
frequency= 1/2pi(g/L)^1/2= so...-->4.02 Hz

vo=2*pi*frequency*amplitude=

v=+/-Vo ( 1-x^2/A^2)^1/2

I was thinking more of;

[tex]v = \sqrt{\frac{T}{\frac{m}{L}}}[/tex]

chazgurl4life said:
i don't know the wave equation

I'm sure you do, [itex]v = f\lambda[/itex]

~H
 
ok so...if i use v= (T/m/l)^1/2
it should be v=(.07154/4.4x10^4 kg/m/.150m)^1/2

which would turn out to be...32.7 m/s...im confused
 
Figure it out, step by step. Answer these questions, in order:
(1) What wavelength must the wave have to exhibit the required number of loops?
(2) What must the speed of the wave be? (Since the frequency is given.)
(3) What string tension is needed to produce such a speed?
(4) What mass will be needed to provide that tension?
 

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