A spring hangs vertically from a fixed support. A mass is then attached to the l
ID: 1778466 • Letter: A
Question
A spring hangs vertically from a fixed support. A mass is then attached to the lower end of a spring. When this system undergoes simple harmonic motion, it has a period of 0.36 s. By how much is the spring stretched from its initial length when the mass and spring are hanging motionless?
A pendulum with a length of precisely 1.000 m can be used to measure the acceleration due to gravity, g. Such a device is called a gravimeter.
A spring hangs vertically from a fixed support. A mass is then attached to the lower end of a spring. When this system undergoes simple harmonic motion, it has a period of 0.36 s. By how much is the spring stretched from its initial length when the mass and spring are hanging motionless?Explanation / Answer
period of Oscilations = T=0.36s
frequency of oscillations = f=1/T=1/0.36=2.78 hz
Angular velocity =w= 2*pi*f=2*pi*2.78=17.46 rad/s
let spring constant be k and mass of block be m
17.46 = sqrt(k/m)
k/m=304.6
m/k=0.003282
extension of sring due to weight of block = mg/k = 0.003282*9.8=0.03217 m = 3.217 cm
a)Length of pendulum = 1m . acceleration due to gravity = 9.81 m/s2
time period of oscillation = 2*pi*sqrt(l/g)=2*pi*(1/9.81) = 2 seconds
Time taken for 114 oscillations = 114*2=224 seconds
b) g=9.81-(0.0012*9.81)=9.798 m/s2
Time period of oscillation = 2*pi*sqrt(1/9.798)=2.007 seconds
Time needed for 114 oscilations = 114*2.007=228.83 seconds
c)yes . the difference can be measured using the stopwatch because the time difference it can measure is less than te actual difference of 0.83 second