Question
A spring with no mass attached to it hangs from a rigid support. A mass m is now hung on the lower end to the spring. The mass is supported on a platform so that the spring remains relaxed. The supporting platformis then suddenly removed and the mass begins to oscillate. The lowest position of the mass during the oscillation is 5 cm below the place where it was resting on the platform. What is the angular frequency of oscillation? Take g = 10 ms−2
Answer: Option B
:
B
It is clear that the separation between the two extreme positions of the oscillatingmass is 5 cm. Therefore, the equilibrium position 0 is 2.5 cm below the supporting platform. Inother words, force mg produces an extension y = 2.5 cm in the spring. If k is the force constantof the spring we have
m g=k y
or km=gy=1000cms−22.5cm=400s−1
The angular frequency ωof oscillation is
ω=√km=√400=20rads−1
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:
B
It is clear that the separation between the two extreme positions of the oscillatingmass is 5 cm. Therefore, the equilibrium position 0 is 2.5 cm below the supporting platform. Inother words, force mg produces an extension y = 2.5 cm in the spring. If k is the force constantof the spring we have
m g=k y
or km=gy=1000cms−22.5cm=400s−1
The angular frequency ωof oscillation is
ω=√km=√400=20rads−1
Was this answer helpful ?
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