#### 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 1

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* = ω^{2} *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/s^{2}

iii) Maximum velocity, *v*_{max} = *A*ω

`= 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.

#### Solution 2

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)`