Question
Download Solution PDFTwo identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from the equilibrium position on either side, it executes a simple harmonic motion. The time period of oscillations of this system is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculations:
Given:
It shows two springs of mass "m" and spring constant '2k' attached to a fixed support.
As we know, Springs connections are the same as resistance circuits. Which gets added simply when they are connected in series.
Total spring constant of this circuit , K = 2k + 2k = 4k
The time period of oscillation is = T = 2π / ω
Where ω = √(K/m) then time period of oscillation = 2π √(m/K) -----(1)
The time period of oscillation is = 2π √(m/K) = \( 2π√{m/4k } \)
Which can be written as T = π \(√{m/k }\) -----(2)
Hence, option (3) is correct.
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