A 800 g body vibrates S.H.M with amplitude 0.30 m. The restoring force is 60 N and the displacement is 0.30 m. Find out (1) T (ii) →a (iii) →v (iv) K.E (v) P.E when the displacement is 12cm.
(i) T = 1.3 s (ii) →a = 3 m s⁻² (iii) →v = 1.4 m s →v (iv) K.E = 7.6 J (v) PE = 1.44 J)


Given:

Mass of the body = m  = 800 g = 0.8 kg

Amplitude = x₀ = 0.30 m

Force (restoring)  = Fr  = 60 N 

Displacement = x = 0.30 m


To Find:

(1) Time Period = T = ? 
(ii) acceleration = →a = ? when the displacement = x = 12cm = 0.12 m
(iii)  speed = →v = ? 
(iv) K.E = ? when the displacement = x = 12cm = 0.12 m
(v) P.E  = ? when the displacement = x = 12cm = 0.12 m


Solution:


(1) Time Period = T = ? by using x = 0.3 m

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

T = 2π √mk

Here we have find first the value of spring constant due to mas of the block m₁. So by using Hook's Law

Fr = k x

or

k = Frx 

by putting values

k = 60N0.30m

k =  200 N m⁻¹ 


Now the Time period is 

T = 2π √mk

by putting values 

T = 2π √0.80kg200Nm⁻¹

T = 2 ｘ 3.1416 ｘ √0.004s²

T = 6.283 ｘ 0.063 s

T = 0.397 s

or

T = 0.4 s ------------------Ans.(i)

( NOTE:  The Answer shown in the book T = 1.3 s can be only obtained if we use Fr  = 6 N thus obtaining K = 20 Nm⁻¹ ).



(ii) acceleration = →a = ? when the displacement = x = 12cm = 0.12 m

According to the Hook's Law

F = k x

and Newton Second law

F = m→a

by comparing we get  

m→a = kx

or

→a = kxm

by putting values

→a = 200Nm⁻¹ｘ 0.12m0.8kg

→a =  30 m s⁻² ---------------Ans(ii)

( NOTE:  The Answer shown in the book a = 3.0 s can be only obtained if we use Fr  = 6 N thus obtaining K = 20 Nm⁻¹ ).



(iii)  speed = →v = ? 

We know that 

v = ധ √x₀2-x2


where ധ = 2πT

by putting values

ധ = 2ｘ3.14161.3s

ധ = 4.8 rad s⁻¹  ----------------Ans(iii)

so, 

v = ധ √x₀2-x2

by putting values

v = 4.8 rad s⁻¹ ｘ √(0.30m)2-(0.12m)2

v = 4.8 rad s⁻¹ ｘ √0.0756m2

v = 4.8 rad s⁻¹ ｘ 0.275

v = 1.319 m s⁻¹ ----------------------Ans. (iii)



(iv) K.E = ? when the displacement = x = 12cm = 0.12 m


The formula for K.E. is

K.E = 12 k (x₀2-x2)

by putting values 

K.E = 12ｘ 200 N m⁻¹ｘ((0.30m)2-(0.12m)2)

K.E =  100 N m⁻¹  ｘ 0.0756m2

K.E = 7.56 J -----------------------Ans (iv)



(v) P.E  = ? when the displacement = x = 12cm = 0.12 m

P.E = 12 k x2

putting values

P.E = 12 200 N m⁻¹ (0.12m)2

P.E = 100 N m⁻¹ ｘ 0.0144 m²

P.E = 1.44 J



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