A loaded spring performs simple harmonic motion with an amplitude of 5cm. If the maximum acceleration of the load is 20cm s-1, calculate the angular frequency of the motion
2 rad s-1
4 rad s-1
5 rad s-1
10 rad s-1
Explanation
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Discussions (5)
To calculate the angular frequency w of the shm,we can use
amax=w^2A
convert cm to mrtres sowe have
0.20=w^2 0.05
w^2=0.20÷0.05
w=√4s^-2
to find frequency
f=w÷2π
f=2s^-1÷2π
so the frequency is 2s^-1÷2π and the frequency is approximately 0.3183Hertz
At maximum acceleration,
Acceleration is equal to the square of the angular frequency multiplied by the amplitude..i.e
a = w²A
Hence, w² = a/A
= ( 20×10^-2 ) / ( 5×10^-2 )
w² = 4
w = 2rad/s
Option A is absolutely right..
At maximum acceleration,
Acceleration is equal to the square of the angular frequency multiplied by the amplitude..i.e
a = w²A
Hence, w² = a/A
= ( 20×10^-2 ) / ( 5×10^-2 )
w² = 4
w = 2rad/s
Option A is absolutely right..
