A load of 15 g elongates a spring by 2 cm. If body of mass 294 g is attached to the spring and is set into vibration with an amplitude of 10 cm, what will be its (i) period (ii) spring constant (iii) maximum speed of its vibration. (Ans: (a) 1.26 s, (b) 7.35 N m⁻¹, (c) 49.0 cm s⁻¹)



Data Given:


Mass of the Load = m₁  = 15 g = 0.015 kg

Elongation due to load = x = 2 cm = 0.02 m

Mass of the body = m₂  = 294 g = 0.294 kg

Elongation due to Body = x = 10 cm = 0.1 m

Value of acceleration due to gravity = g = 9.8 m s⁻²



To Find:


(a) Tine Period of the vibration due to body mass m₂  = T = ?

(b) Spring constant = K = ?

(c) Maximum Speed of body vibration = v₀ = ?


Solution:


The general formula for Time Period of a mass attached to spring is 

T = 2Ï€ `\sqrt frac{m}{k}`

Here we have find the value of spring constant first from the given for mass of the block m₁. 

(b) Spring constant = K = ?


By using Hook's Law

F = k x

or

k = `\frac {F}{x}` = `\frac {mg}{x}`

by putting values

k = `\frac {0.015 kg ï½˜ 9.8 m s⁻²}{0.02 m}`

k =  7.35 N m⁻¹  --------------Ans.(b)


(a) Tine Period of the vibration due to body mass m₂  = T = ?

Now due to the vibration of body of mass m₂, the Time period is 

T = 2Ï€ `\sqrt frac{m₂}{k}`

by putting values 

T = 2Ï€ `\sqrt frac{0.294 kg}{7.35 N m⁻¹}`

T = 2 ï½˜ 3.1416 ｘ `sqrt 0.04 s²`

T = 6.292 ï½˜ 0.2 s

T = 1.2584 s

T = 1.26 s ------------------Ans. (a)


(c) Maximum Speed of body vibration = v₀ = ?

Using equation of maximum velocity


v₀ = x ( `\sqrt frac{k}{m₂}`)

by putting values

v₀ = 0.1 m (`\sqrt frac{7.35 N m⁻¹}{0.294 kg}`)

v₀ = 0.1 m (`sqrt 25 s⁻²`)

v₀ = 0.1 m ｘ 5 s⁻¹

v₀ = 0.5 m s⁻¹ ----------------------Ans. (c)

or

v₀ = 50 cm s⁻¹ ----------------------Ans. (c)




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