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An undamped spring-mass system with mass m and spring stiffness k is shown in the figure. The natural frequency and natural period of this system are ω rad/s and T s, respectively. If the stiffness of the spring is doubled and the mass is halved, then the natural frequency and the natural period of the modified system, respectively, are

An undamped spring-mass system with mass m and spring stiffness k is shown in the figure. The natural frequency and natural period of this system are ω rad/s and T s, respectively. If the stiffness of the spring is doubled and the mass is halved, then the natural frequency and the natural period of the modified system, respectively, are Correct Answer 2ω rad/s and T/2 s

Concept

Natural frequency (Wn) of any undamped single degree of freedom system is given by

Wn = √(k/m) 

Wn in rad/sec

k = Stiffness of system in N/m

m = mass of system in kg

Time period (T) = 2π / Wn

Given data and Calculation-

Given that, Initial mass (m0) = m and stiffness (k0) = k

Natural frequency = ω rad/s 

Natural time period = T s

Final mass m1 = m/2 and Stiffness k1 = 2k

Then, Wn1 = √(k1/m1) = √(4k/m) = 2 √(k/m) = 2 Wn  

T1 = 2π/Wn1 = 2π/(2 Wn) = T/2 

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