In figure (a), string 1 has a linear density of 3.20 g/m, and string 2 has a lin
ID: 1907223 • Letter: I
Question
In figure (a), string 1 has a linear density of 3.20 g/m, and string 2 has a linear density of 4.10 g/m. They are under tension due to the hanging block of mass M = 500 g.
(a) Calculate the wave speed on string 1. Incorrect: Your answer is incorrect. m/s
(b) Calculate the wave speed on string 2. m/s
The block is now divided into two blocks (with M1 + M2 = M) and the apparatus is rearranged as shown in figure (b). Find M1 and M2 such that the wave speeds in the two strings are equal.
(c) M1 g
(d) M2 g
Explanation / Answer
a)v=T/m=500*10^-3*9.8/3.2*10^-3=39.13m/s
b) wave does not pass into the string 2 as there ia a knot there so v=0
v1=v2
T1/m1=T2/m2
T1/m1=T2/m2
M1g/m1=M2g/m2
M1/m1=M2/m2
M1/M2=m1/m2=3.2/4.1
M1=0.78M2
M1+M2=500g
0.78M2+M2=500
M2=500/1.78=280.89g
M1=219.11g