Question

In SI units, the differential equation of an S.H.m is =-16x. Find its frequency and period.
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Answer 1

Answer:

t=(2*3.14)/4=3.14/2=1.57 sec

Explanation:

Simple harmonic motion is defined by the equation:a=(-x)w^2 ——(1)

where a and x have usual meaning as in question and w=angular frequency .

The relation between time period(t) and angular frequency is:: t=2pi/w ——(2)

By comparing a=-16x and eq(1) we get w=4

From eq(2) time period t=(2*3.14)/4=3.14/2=1.57 sec

Answer 2

The differential equation of a simple harmonic motion (SHM) is given by:

d²x/dt² = -kx

Where x is the displacement from the equilibrium position, t is time, and k is the spring constant.

For a mass-spring system, the value of k is given by:

k = mω²

Where m is the mass of the object, and ω is the angular frequency.

Comparing the given differential equation with the general form of SHM equation, we can see that k = 16.

So, k = mω² = 16

ω² = k/m = 16/m

ω = sqrt(16/m) = 4sqrt(m)

The frequency of the motion is given by:

f = ω/2π

Substituting the value of ω, we get:

f = 4sqrt(m)/2π = 2sqrt(m)/π

The period of the motion is given by:

T = 1/f

Substituting the value of f, we get:

T = 1/(2sqrt(m)/π) = π/(2sqrt(m))

Therefore, the frequency of the SHM is 2sqrt(m)/π and the period is π/(2sqrt(m)) in SI units. Note that the mass, m, is not specified in the problem, so we cannot give numerical values for the frequency and period without more information.

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