A string is attached to the blade at one end, passes over a fixed pulley and is attached to a hanging block of mass M at the other end. Two cases are shown. In case "a," the block hangs vertically down and the string vibrates in the second harmonic. In case "b," a fan blows the block to the right by applying a constant horizontal force F and the string vibrates in the fundamental. Determine the magnitude of the force the fan must exert on the block in case "b." (Use any variable or symbol stated above along with the following as necessary: g.) The figure below shows a vibrating blade at the left which maintains a constant frequency.MaFーMbA string is attached to the blade at one end, passes over a fixed pulley and is attached to a hanging block of mass M at the other end. Two cases areshown. In case "a," the block hangs vertically down and the string vibrates in the second harmonic. In case "b," a fan blows the block to the right byapplying a constant horizontal force F and the string vibrates in the fundamental. Determine the magnitude of the force the fan must exert on theblock in case "b." (Use any variable or symbol stated above along with the following as necessary: g.)F =

The block of mass M suspended from the pulley exerts a force of tension on the string. This force of tension on the string is balanced by the gravitational force exerted on the mass M Now, as the string is vibrating, a standing wave of certain velocity is produced over the string. In the case a, a second harmonic wave of velocity v is produced on the string. Since no other force is being exerted on the string other than the gravitational force of the mass M Now, as a horizontal force F is acted on the mass M, there is a shift in the mass M from its original position. Now the mass M makes an angle θ with the vertical axis downwards. Thus, due to this, the gravitational force can be resolved into two components. The first component Mgcosθ, along downward direction, and the second component Mgsinθ along the horizontal direction. The component Mgcosθ of the gravitational force is balanced by the force of tension in the string. Now, the horizontal force F is along the same direction as the component Mgsinθ, and it should also have magnitude equal to the magnitude of the horizontal component of the gravitational force So, F=Mgsinθ Answer: The horizontal force must have magnitude equal to the horizontal component of the gravitational force F=Mgsinθ

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Advanced Physics

A string is attached to the blade at one end, passes over a fixed pulley and is attached to a hanging block of mass M at the other end. Two cases are shown. In case "a," the block hangs vertically down and the string vibrates in the second harmonic. In case "b," a fan blows the block to the right by applying a constant horizontal force F and the string vibrates in the fundamental. Determine the magnitude of the force the fan must exert on the block in case "b." (Use any variable or symbol stated above along with the following as necessary: g.)

A string is attached to the blade at one end, passes over a fixed pulley and is attached to a hanging block of mass M at the other end. Two cases are shown. In case "a," the block hangs vertically down and the string vibrates in the second harmonic. In case "b," a fan blows the block to the right by applying a constant horizontal force F and the string vibrates in the fundamental. Determine the magnitude of the force the fan must exert on the block in case "b." (Use any variable or symbol stated above along with the following as necessary: g.)

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