wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A simple pendulum of length l having a bob of mass m is suspended in a car that is travelling with a constant speed v around a circular path of radius R. If the pendulum undergoes small oscillations about its equilibrium position, the frequency of its oscillation will be

A
12πgl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
12π   (g2+v4R2)12l
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
12πgR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12πv2Rl
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 12π   (g2+v4R2)12l
Time period of simple pendulum in non-inertial frame is T=2πlgeff
Due to circular motion, centripetal acceleration on the bob is =v2R towards the centre.
Hence,
geff=g2+(v2R)2

Time period is
T=2π   l(g2+v4R2)12
Frequency of oscillation
f=1T=12π   (g2+v4R2)12l

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon