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Question

Physics Question on Oscillations

A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 13.19. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm and released.

Determine (i) the frequency of oscillations, (ii) maximum acceleration of the mass, and (iii) the maximum speed of the mass

Answer

Spring constant, k = 1200 N m-1

Mass, m = 3 kg

Displacement, A = 2.0 cm = 0.02 cm

Frequency of oscillation v, is given by the relation:

v=1T=12πkmv=\frac{1}{T}=\frac{1}{2\pi}\sqrt\frac{k}{m}

Where, T is the time period

v=12x3.1412003=3.18m/s∴v=\frac{1}{2x3.14}\sqrt\frac{1200}{3}=3.18\,m/s

Hence, the frequency of oscillations is 3.18 m/s

Maximum acceleration (a) is given by the relation:

a = ω2 A

Where,

ω = Angular frequency = km\sqrt\frac{k}{m}

A = Maximum displacement

a=kmA=1200×0.023=8ms2\therefore\,a=\frac{k}{m}A=\frac{1200×0.02}{3}=8\,ms^{-2}

Hence, the maximum acceleration of the mass is 8.0 m/s2

Maximum velocity, v max = Aω

=Akm=0.02x12003=0.4m/s.=A\sqrt\frac{k}{m}=0.02x\sqrt\frac{1200}{3}=0.4 \,m/s.

Hence, the maximum velocity of the mass is 0.4 m/s.