A block is attached to a horizontal spring and oscillates back and forth on a fr

Question

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of 2.91 Hz. The amplitude of the motion is 5.60 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 the frequency of the simple harmonic motion that exists after the block splits? amplitude = m frequency = Hz (b) Repeat part (a), assuming that the block splits when it is at one of its extreme positions. amplitude = m frequency = Hz

Explanation / Answer

initial angular frequency, w = 2*pi*f

= 2*pi*2.91

= 18.28 rad/s

Let k is spring constant.

iniial Amplitude, A = 0.0560 m

a) let A ' is the amplitude of motion when half of the mass is gone.

0.5*k*A'^2 = 0.5*(m/2)*vmax^2

0.5*k*A'^2 = 0.5*(m/2)*(A*w)^2

0.5*k*A'^2 = 0.5*(m/2)*w^2*A^2

k*A'^2 = (m/2)*(k/m)*A^2

A'^2 = A^2/2

A' = A/sqrt(2)

= 0.0560/sqrt(2)

= 0.0395 m or 3.95*10^-2 m

b) f' = sqrt(k/(m/2))/(2*pi)

= sqrt(2)*sqrt(k/m)/(2*pi)

= f*sqrt(2)

= 2.91*sqrt(2)

= 4.11 hz

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