Object A is attached to ideal spring A and is moving in simple harmonic motion.

Question

Object A is attached to ideal spring A and is moving in simple harmonic motion. Object B is attached to ideal spring B and is moving in simple harmonic motion. The period and the amplitude of object B are both two times the corresponding values for object A. How do the maximum speeds of the two objects compare?

a) The maximum speed of A is one fourth that of object B. b) The maximum speed of A is one half that of object B.
c) The maximum speed of A is the same as that of object B. d) The maximum speed of A is two times that of object B. e) The maximum speed of A is four times that of object B.

Explanation / Answer

c) The maximum speed of A is the same as that of object B. is the answer

Given

Period

TB=2TA

amplitude

AB=2AA

Let us calculate the Maximum Speed of object A is

Vmax,A=AW=AA*(2pi/TA)

maximum speed of B is

Vmax,B =AW =A*(2pi/T)=(2AA) *(2pi/2TA)

Vmax,B=AA*(2pi/TA)=Vmax,A

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.