Answer:a) T=0.44seconds and f=2.29Hz
b) A=0.1626m
c) amax=33.11m/s^2
d) E=2.2337J
e) KE=1.8732J
Step-by-step explanation:
Period of the motion :
T=2π√m/k
T=period,m=mass of the body, k=spring constant
Converting mass from gram to kilogram :
M=830g/1000=0.83kg
T=2π√0.83/169=0.44038seconds
T=0.44secs(approximately)
Frequency of the motion :
F=1/T
Where T=0.44038seconds
F=1/0.44038s=2.27Hz
b) maximum speed :
Vmax=A√k/m
Where A=amplitude,m=mass of the body, k=the spring constant
Making A the subject of the formula:
A=Vmax√m/k
Where, speed(V)=2.32m/s as Vmax and 0.83kg for m and 169N/m for k.
A=(2.32m/s)√(0.83kg)/(169N/m)= 0.1626m
c) Maximum acceleration :
amax =A(k/m)
Therefore, A=
(0.1626m)(169N/m)/0.83kg)
A=33.11m/s^2
d) Total Energy :
E=1/2MVmax
Where, M=0.83kg and Vmax=2.32m/s
E=1/2*(0.83kg)*(2.32m/s)^2
E=2.2337J
e) From the law of conservation of mass :
1/2kA^2=1/2Kx^2+KE
Where x=0.04A
KE=1/2kA^2-1/2kx^2
KE=1/2kA^2(1-0.4^2)
KE=Etotal*0.84
Where, Etotal=2.23J
KE=2.23*0.84=1.8732J