1. May 5, 2006

chazgurl4life

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?

2. May 5, 2006

Hootenanny

Staff Emeritus
Hi chazgurl, you should have realised 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

3. May 7, 2006

chazgurl4life

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 dont know the wave equation

4. May 7, 2006

Hootenanny

Staff Emeritus
I was thinking more of;

$$v = \sqrt{\frac{T}{\frac{m}{L}}}$$

I'm sure you do, $v = f\lambda$

~H

5. May 8, 2006

chazgurl4life

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

6. May 8, 2006

Staff: Mentor

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?