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 = 2π √mk√mk

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 = FxFx = mgxmgx

by putting values

k = 0.015kg ｘ9.8ms⁻²0.02m

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 = 2π √m₂k

by putting values 

T = 2π √0.294 kg7.35Nm⁻¹

T = 2 ｘ 3.1416 ｘ √0.04s²

T = 6.292 ｘ 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 ( √km₂)

by putting values

v₀ = 0.1 m (√7.35Nm⁻¹0.294kg)

v₀ = 0.1 m (√25s⁻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)




************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.