In mass spring system mass 'm' is attached with spring of spring constant 'k' with time period 'T1'..Then the mass is replaced by '2m' with same spring, what is the time period 'T2'
Correct answer: C. T2=β2T1
- A. T2=T1
- B. T2=2T1
- C. T2=β2T1
- D. T2=T1/(β2)
Explanation
Time period T for a mass attached to a spring can be calculated by using the following formula: T=2π βm/k T1=2π βm/k ==============eq(1)Since 2m is attached to the same spring to which m was attached, spring constant k remains same. T2=2π β2m/k ==============eq(2)Divide eq(2) by eq(1)T2/T1=2π β2m/k / 2π βm/kT2=β2xT1So increasing the mass by 2 would increase the time period by β2Correct answer here is option C
This question appeared on the UHS MDCAT 2017 paper, which you can sit online with every answer explained.
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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