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.

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

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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