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GATE 2022 CE (CE2) – Question 37

Engineering Mechanics · Internal forces in structures · 2 marks · Multiple choice

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

Mass m attached to a horizontal spring of stiffness k. See the attached source image.
  1. $2\omega,T/2$
  2. $\omega/2,2T$
  3. $4\omega,T/4$
  4. $\omega,T$

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Show answer and explanation

Correct answer: (A) $2\omega,T/2$

Explanation

Natural frequency is $\omega=\sqrt{k/m}$ and period is $T=2\pi/\omega$. The modified ratio is $\sqrt{(2k)/(m/2)}/\sqrt{k/m}=2$.

Thus frequency doubles and period halves: **2ω and T/2 (A)**.