the time period of a simple pendulum is T if its point of suspension is moved upward according to relation y =
Answer (1)
Hi,
We know that time period of pendulum ofllength with net vertical accelerationanetis given by
y=kt^2
=d^2y/dt^2=2k
a=2m/s^2
T1=2pie under root(g/l)
T2=2pie under root(l/g+ay)
(T1/T2)^2=(g+ay)/g
Using the values we get that
(10+2)/10=6/5=1.2
We know that time period of pendulum ofllength with net vertical accelerationanetis given by
y=kt^2
=d^2y/dt^2=2k
a=2m/s^2
T1=2pie under root(g/l)
T2=2pie under root(l/g+ay)
(T1/T2)^2=(g+ay)/g
Using the values we get that
(10+2)/10=6/5=1.2
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