A spring having with a spring constant 1200 N m–1 is mounted on a horizontal table as shown in Fig. 14.24. A mass of 3 kgis 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.
Given: The spring constant of spring is 1200 N / m , the mass attached at free end is 3 kg , and the mass is pulled by a distance of 2 cm .
(i)
The frequency of oscillation is given as,
ν = 1 2 π k m
Where, the frequency of oscillation is ν , the mass is m and the spring constant is k .
By substituting the given values in above equation, we get
ν = 1 2 π 1200 3 = 1 2 π 400 = 3.18 Hz
Thus, the frequency of oscillation is 3.18 cycle per second .
(ii)
The angular frequency of spring is given as,
ω = k m
The magnitude of maximum acceleration is given by,
a = ω 2 A = k m A
By substituting the values in above equation, we get
a = 1200 3 × 0.02 = 8 m / s 2
Thus, the maximum acceleration of the mass is 8 m / s 2 .
(iii)
The maximum velocity is given by,
v = A ω = A × k m
By substituting the values in above equation, we get
v = 0.02 × 1200 3 = 0.02 × 20 = 0.4 m / s
Thus, the maximum velocity of the mass is 0.4 m / s
Given: The spring constant of spring is
(i)
The frequency of oscillation is given as,
Where, the frequency of oscillation is
By substituting the given values in above equation, we get
Thus, the frequency of oscillation is
(ii)
The angular frequency of spring is given as,
The magnitude of maximum acceleration is given by,
By substituting the values in above equation, we get
Thus, the maximum acceleration of the mass is
(iii)
The maximum velocity is given by,
By substituting the values in above equation, we get
Thus, the maximum velocity of the mass is
