A 1 kg block attached to a spring vibrates with a frequency of 1 Hz on a frictionless horizontal table. Two springs identical to the original spring are attached in parallel to an 8 kg block placed on the same table. What is the frequency of vibration of the 8 kg block?
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The frequency of oscillation of a mass m when set up as a harmonic oscillator with a spring has a frequency of oscillation F = (1/2*pi)*sqrt(k/m).
The 1 kg block attached to a spring vibrates with a frequency of 1 Hz. This gives the spring constant of the spring as k where (1/2*pi)*sqrt(k/1) = 1
taking the square of both the sides and isolating k
k = 1*4*pi^2*1
=> k = 4*pi^2 N/m
When two of these springs are attached in parallel the resulting spring constant is equal to Keq = 4*pi^2 + 4*pi^2
=> Keq = 8*pi^2
The frequency of vibration of an 8 kg block that is attached to the springs and placed on a frictionless surface is (1/2*pi)*sqrt(8*pi^2/8)
=> (1/2*pi)*(pi)
=> 1/2
The frequency of vibration of the 8 kg block is 1/2 Hz.
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