Question

Review Conceptual Example 8 before starting this problem. A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 4.67 Hz. The amplitude of the motion is 8.53 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring. (a) What is the amplitude and (b) the frequency of the simple harmonic motion that exists after the block splits?

Answer 1

original Amplitude, A0 = 8.53 x 10^-2 m
original frequency, f₀= 4.67 Hz.
original mass of block = M
Let new Amplitude and freq be A₁ and f₁ respectively.

all the energy is in kinetic form inside blocks. Half of this energy is carried away with free block and our new system will be having only half of the initial energy.

(a)
0.5 x k x A₁²= 1/2(0.5 x k x A₀²) =
=> A₁= A₀(1/2) = 8.53 x 10^-2 / 1.414 = 6.03 x 10^-2 m
(b)
mass of attached block = M/2, hence
f1= (1/2) k / (M/2) =   2 x f₀ = 1.414 x 4.67 = 6.60 Hz

answered by: anonymous