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प्रश्न
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.
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उत्तर १
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:
`v = 1/T = 1/(2pi) sqrt(k/m)`
Where, T is the time period
`:. v = 1/(2xx3.14) sqrt(1200/3) = 3.18 "m/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
ω = Angular frequency = `sqrt(k/m)`
A = Maximum displacement
`:. a = k/m A = (1200xx0.02)/(3) = 8 ms^(-2)`
Hence, the maximum acceleration of the mass is 8.0 m/s2
iii) Maximum velocity, vmax = Aω
`= A sqrt(k/m) = 0.02 xx sqrt(1200/3) = 0.4 "m/s"`
Hence, the maximum velocity of the mass is 0.4 m/s.
उत्तर २
K = 1200 `Mn^(-1)`; m = 3.0 kg, a= 2.0 cm = 0.02 m
i) Frequency, `v = 1/T = 1/(2pi) sqrt(k/m) = 1/(2xx3.14) sqrt(1200/3) = 3.2 s^(-1)`
ii) Acceleration, A = `omega^2` `" " y = k/m y`
Acceleration will be maximum when y is maximum i.e y = q
:. max acceleration,` A_"max" = (ka)/m =(1200xx0.02)/3 = 8 ms^(-2)`
iii) Max speed of the mass will be when it is passing throught mean position
`V_"max" = aomega = sqrt(k/m) = 0.02 xx sqrt(1200/3) = 0.4 ms^(-1)`
संबंधित प्रश्न
let us take the position of mass when the spring is unstretched as x = 0, and the direction from left to right as the positive direction of the x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t = 0), the mass is
(a) at the mean position,
(b) at the maximum stretched position, and
(c) at the maximum compressed position.
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