Question

A mass of 2 kg is suspended from a steel wire of length 1 meter to form a pendulum arrangement. Of the mass is moved to one side and released from the horizontal position of wire then find the maximum extension in the length of the wire.

Answer 1

A mass of 2 kg is suspended from a steel wire of length 1 meter to form a

pendulum arrangement.

Of the mass is moved to one side and released from the horizontal position of wire then

find the maximum extension in the length of the wire.?

given Ysteel=2 x 10^11 N/m^2 area of cross

 section of wire =1  mm^2 ,g=10m/s^2

velocity at lowest position =√2g=√2*10=√20 m/s        ,∴g=10m/s^2

now force at lowest position F=2g+2*20/1=  2*10+40  = 20+40    = 60 N

Y=60/10^-6/ΔL/1  

2×10^11=60/ΔL*10^-6

60/2*×10^11*10^-6=ΔL

ΔL=3*10^-4 m

ΔL=0.3 mm

maximum extension in the length of wire is  0.3mm

.

Answer 2

Answer:

Explanation:

*See the attachment below*