Question

A spring having with a spring constant 1200 N m^–1 is mounted on a horizontal table as shown in Fig. 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.

Solution

Spring constant, k = 1200 N m^–1
Mass, m = 3 kg
Displacement, A = 2.0 cm = 0.02 cm
(i) Frequency of oscillation v , is given by the relation:

straight nu space equals space 1 over straight T space equals space fraction numerator 1 over denominator 2 straight pi end fraction square root of straight k over straight m end root where comma space straight T space is space the space time space period. space Therefore comma space straight nu space equals space fraction numerator 1 over denominator 2 space straight x 3.14 end fraction space square root of 1200 over 3 end root space space space space equals space 3.18 space straight m divided by straight s

Hence, the frequency of oscillations is 3.18 cycles per second.
(ii) Maximum acceleration (a) is given by the relation,
a = ω² A

where,

straight omega space equals space Angular space frequency space equals space square root of straight k over straight m end root straight A space equals space maximum space displacement therefore space straight a space equals space straight k over straight m space straight A space space space space space space space space space space equals space fraction numerator 1200 space straight x space 0.02 over denominator 3 end fraction space space space space space space space space space space equals space 8 space straight m divided by straight s squared

Hence, the maximum acceleration of the mass is 8.0 m/s².
(iii) Maximum velocity, v_max = Aω

straight A space square root of straight k over straight m end root space equals space 0.02 space straight x space square root of 1200 over 3 end root space space space space space space space space space space space space space space space space space space equals space 0.4 space straight m divided by straight s

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

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