A spring having with a spring constant \( 1200 N m^{-1}\) is mounted on a horizontal table as shown in given figure .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.

A spring having with a spring constant 1200 N m ^-1 is mounted on a  horizontal table as shown in the Figure. A mass of 3 kg is attached to the  free

Determine

(i) the frequency of oscillations,

(ii) maximum acceleration of the mass, and

(iii) the maximum speed of the mass.

Asked by Abhisek | 1 year ago |  120

1 Answer

Solution :-

Given,

Spring constant, k = 1200 Nm-1

Mass, m = 3 kg

Displacement, A = 2.0 cm

= 0.02 m

(i) Frequency of oscillation ‘v’ is given by the relation:

v = \( \dfrac{1}{T}\)

= (\( \dfrac{1}{2 π}\)) (\( \dfrac{\sqrt{k}}{m}\))

Where,

T is the time period

So,

v = \(( \dfrac{1}{2 \times 3.14})\)\( \dfrac{\sqrt{1200}}{3}\)

On calculating further, we get,

= 3.18 m/s

Therefore, 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 = \( \dfrac{\sqrt{k}}{m}\)

A = Maximum displacement

Hence,

a = \( \dfrac{K}{m}A\)

a = \( \dfrac{(1200 \times 0.02)}{3}\)

We get,

= 8 ms-2

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

(iii) Maximum velocity, vmax = Aω

On substituting, we get,

= \( \dfrac{A\sqrt{k}}{m}\)

= 0.02 x (\( \dfrac{\sqrt{1200}}{3}\))

= 0.4 m/s

Therefore, the maximum velocity of the mass is 0.4 m/s

Answered by Abhisek | 1 year ago

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