derive the expression for the time period of oscillation of spring T=2pi rootof m÷k, where T is time period, m is mass of the body and K is force constant?
Answers
Answer :
Step by step explanation :
There's one more simple method for deriving the time period. The spring pendulum, as we all know is a great example for Simple Harmonic Motion. First, let's assume a particle at any point of the spring. While in an SHM, the acceleration of the particle (at some position on the pendulum) is directly proportional to its displacement (x), and is always directed towards the fixed point of the spring. From the past observations,
a α x⟹a=−ω2x
I think you know that ω is the angular frequency of this particle and the negative sign indicates that the acceleration is opposite to the direction of displacement. If the mass is given by m, then the restoring force is given by substituting the above equation in Newton's second law. (For instance, it is F=ma)
I think you know that ω is the angular frequency of this particle and the negative sign indicates that the acceleration is opposite to the direction of displacement. If the mass is given by m, then the restoring force is given by substituting the above equation in Newton's second law. (For instance, it is F=ma)The restoring force,
F=−mω2x
Here, the mω2 is substituted by k. Its always a constant for a given spring (and hence the name, spring constant). From this, we could simply get
ω=√m ÷ √k
Now, time period T is the time taken by the spring to complete one oscillation with a given ω. We could define angular velocity using this time period. A complete revolution per T, gives the angular frequency.
ω=2πT
Equating both, you obtain
T=2π√m
√k
Explanation:
(a) For a simple pendulum, force constant or spring factor k is proportional to mass m, therefore, m cancels out in denominator as well as in numerator. That is why the time period of simple pendulum is independent of the mass of the bob.
(b) In the case of a simple pendulum, the restoring force acting on the bob of the pendulum is given as:
F=–mgsinθ
where,
F= Restoring force
m= Mass of the bob
g= Acceleration due to gravity
θ= Angle of displacement
For small θ,sinθ≃θ
For large θ, sinθ is greater than θ.
This decreases the effective value of g.
Hence, the time period increases as:
T=2π
g
l
(c) Yes, because the working of the wrist watch depends on spring action and it has nothing to do with gravity.
(d) Gravity disappears for a object under free fall, so frequency is zero.
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Assertion
The amplitude of an oscillating pendulum decreases gradually with time
Reason
The frequency of the pendulum decreases with time.
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If a of simple pendulum having negative charge q