If a system is displaced from its equilibrium position and released, it moves according to the equation ¨θ=−I2klθ where I, k and l are constants. It will oscillate with a frequency:
√I2kl
12π√klI2
2π√klI2
12π√I2kl
A
√I2kl
B
12π√klI2
C
2π√klI2
D
12π√I2kl
Open in App
Solution
Verified by Toppr
Comparing the equation, d2xdt=−ω2x with d2θdt=−I2klθ, we find that ω=√I2kl Now frequency =1T=ω2π=12π×√I2kl. Hence, option C is correct option.
Was this answer helpful?
0
Similar Questions
Q1
If a system is displaced from its equilibrium position and released, it moves according to the equation ¨θ=−I2klθ where I, k and l are constants. It will oscillate with a frequency:
View Solution
Q2
2KI+I2+22HNO3→2HIO3+2KIO3+22NO2 If 3 mole of Kl & 2 moles I2, are reacted with excess of HNO3.
What is the volume of NO2 gas evolved at NTP?
View Solution
Q3
One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of cross – section and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period given by
View Solution
Q4
Figure shows a block of mass m attached to a spring of force constant k and connected to ground by two string. In relaxed state natural length of the spring is l. In the situation shown in figure, find the tension in the strings (1) and (2).
View Solution
Q5
Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from its equilibrium position and released, find the period of its vertical oscillation.