Oscillations

JEE Main level, single correct, medium

104 of 194

Question 104 of 194

A spring stretches by 22 mm when it is loaded with a mass of 200200 g. From equilibrium position the mass is further pulled down by 22 mm and released. The frequency associated with the system and maximum energy in the spring are __________ Hz and __________ J, respectively. (Take g =10= 10 m/s2^2)

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Answer: (A) 550π\dfrac{5\sqrt{50}}{\pi} and 8×10−38 \times 10^{-3}

At equilibrium, the restoring force of the spring balances the weight of the mass:

mg=kx0mg = kx_0

k=mgx0=0.2×102×10−3=1000k = \dfrac{mg}{x_0} = \dfrac{0.2 \times 10}{2 \times 10^{-3}} = 1000 N/m

The frequency of oscillation is given by:

f=12πkmf = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}

f=12π10000.2=12π5000=10502π=550πf = \dfrac{1}{2\pi} \sqrt{\dfrac{1000}{0.2}} = \dfrac{1}{2\pi} \sqrt{5000} = \dfrac{10\sqrt{50}}{2\pi} = \dfrac{5\sqrt{50}}{\pi} Hz

The maximum energy in the spring corresponds to its maximum elastic potential energy, which occurs at the maximum extension.

Maximum extension xmax=x0+A=2 mm+2 mm=4 mm=4×10−3x_{max} = x_0 + A = 2 \text{ mm} + 2 \text{ mm} = 4 \text{ mm} = 4 \times 10^{-3} m

Maximum energy in the spring Umax=12kxmax2U_{max} = \dfrac{1}{2} k x_{max}^2

Umax=12×1000×(4×10−3)2U_{max} = \dfrac{1}{2} \times 1000 \times (4 \times 10^{-3})^2

Umax=500×16×10−6=8000×10−6=8×10−3U_{max} = 500 \times 16 \times 10^{-6} = 8000 \times 10^{-6} = 8 \times 10^{-3} J

Answer: 550π\dfrac{5\sqrt{50}}{\pi} and 8×10−38 \times 10^{-3}

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