Two masses are connected via a massless spring whose lenght is 2m. One mass (m1=
ID: 1417743 • Letter: T
Question
Two masses are connected via a massless spring whose lenght is 2m. One mass (m1=1kg) is spining on a frictionless table top in a cirrcular motion while the other (m2=0.5kg) is suspended 0.5m below the table; what is the perioid of the m1, the top mass, such that m2 remains stationary? Two masses are connected via a massless spring whose lenght is 2m. One mass (m1=1kg) is spining on a frictionless table top in a cirrcular motion while the other (m2=0.5kg) is suspended 0.5m below the table; what is the perioid of the m1, the top mass, such that m2 remains stationary?Explanation / Answer
In the case of a spring. the force can be calculated as F=-kx, where F is the restoring force, k is the force constant, and x is the displacement. The motion of a mass on a spring can be described as Simple Harmonic Motion (SHM): oscillatory motion that follows Hooke's Law.
in a spring period T = 2••(m/k)0.5
period of M1 T=2**(0.5)0.5 = 4.443 sec
The period depends on the length of the swing and also to a slight degree on the amplitude, the width of the pendulum's swing.
period of M2 T=(2*)(L /g)0.5 =(2*pi)(0.5/g)0.5 =1.4185 sec
If m2 remains stationary then T for M2 is zeros and T for M1 remains the same
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.