# Science

With the angle shown as 30º and the inclined plane and pulley being frictionless, the string supports a mass M at the bottom of the plane. The system is in equilibrium. Standing waves are set up on the VERTICAL section of the system.

• find the tension in the string

• the whole length of the string ignoring the radius of the pulley

• the mass per unit length of the string

• the speed of the waves on the string

• the lowest frequency standing wave

• the period of the standing wave having 3 nodes

• the wavelength of the standing waves having 3 nodes

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This image has been Flagged as inappropriate Click to unflag 1) Since the system is in equilibrium, the vector sum of the forces acting on the block is 0. The forces acting on the block are gravity, force from the string (with the magnitude equal to the tension in the string, T), and the normal force. The normal force is...

1) Since the system is in equilibrium, the vector sum of the forces acting on the block is 0. The forces acting on the block are gravity, force from the string (with the magnitude equal to the tension in the string, T), and the normal force. The normal force is perpendicular to the incline and pointing up and is balanced out by the component of the gravity perpendicular to the incline, pointing down. The force from the string on the block is balanced out by the component of the gravity parallel to the incline:

`T = mgsin(theta)`

So the tension in the string equals `mgcos(theta)` .

2) The length of the part of the string along the incline, l,  can be found from the right triangle formed by the incline and the horizontal. Since `sin(theta)` = opposite/hypothenuse,

`sin(theta) = h/l`

From here `l = h/sin(theta)`

Then the length of the whole string is `L =h+h/sin(theta)` .

3) The mass per unit length is

`m/L = m/(h+h/sin(theta)) = (msin(theta))/(hsin(theta) + h)`

4) The speed of the waves in the string is given by

`v = sqrt(T/(m/L))`

Plugging in the mass per unit length and tension results in

`v = sqrt((mgsin(theta))/((msin(theta))/(hsin(theta) + h)))`

`v = sqrt(gh(sin(theta) + 1))`

5) Since the standing wave is set up only in the vertical part of the system, the lowest frequency standing wave will be

`f = v/(2h) = sqrt(gh(sin(theta) + 1))/(2h)`

`f = 1/2sqrt(g/h (sin(theta) + 1))`

6) If the standing wave has 3 nodes, its frequency is 4 times greater, so it will be

`2sqrt(g/h (sin(theta) + 1))`

Then, the period of this wave is `T = 1/f = 1/2sqrt(h/(g(sin(theta) +1)))`

7) The wavelength of this wave will be `lambda = vT` , or

`lambda = sqrt(gh(sin(theta)+1))*1/2sqrt(h/(g(sin(theta)+1)))`

`lambda = h/2`

The wavelength would be h/2, as expected for the standing wave with 3 nodes in the string of length h.

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