Calculating Mass and Wave Speed in a String System

AI Thread Summary
The wave speed in the string system is 26 m/s with a suspended mass of 3.2 kg. To find the mass per unit length of the string, the tension due to the mass is used in calculations. When the suspended mass is reduced to 2.5 kg, the wave speed must be recalculated based on the new tension. The discussion seeks assistance in determining these values. Accurate calculations are essential for understanding wave dynamics in string systems.
jh24628
Messages
23
Reaction score
0
tension is maintained in a string as in the figure. the observered wave speed is 26 m/s when the suspended mass is 3.2 kg

what is the mass per unit length of the string in kg/m

what is the wave speed of the suspended mass when it is 2.5 kg. ans in m/s

please help

(the picture has a string attached to a vertical wall then a pulley and then the mass hanging from the pulley)
 
Physics news on Phys.org
cancel
i got it
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top