Question

The motion of a simple pendulum is approximately simple harmonic for small angle oscillations. For large angles of oscillation, a more involved analysis shows that T is greater than $2\pi \sqrt{\frac {I}{g}}$. Think of a quantitative argument to appreciate this result.

Answer 1

In case of a simple pendulum, the restoring force which is causing simple harmonic motion of bob of pendulum is
F = -mgSin$\theta$
Where m is the mass of the Bob, g is acceleration due to gravity and  is the angle made by string of pendulum with the vertical.

We know force on a Body is
F = ma
Where m is the mass of particle and a is the acceleration of the particle, so acceleration of the body can be written as
a = F/m
so acceleration of bob is
a = -mgSin$\theta$/m = -gSin$\theta$
here we approximate for small oscillations
Sin$\theta$≈$\theta$
But for larger angles  > sin
Further we express displacement as
x = l$\theta$
where x is the displacement of bob making an circular arc and  is the angle covered and l is the length of pendulum so we get
$\theta$ = x/l
And acceleration of pendulum as
a = -(g/l)x
now without using approximation
we get angular frequency
$\omega<\sqrt{\frac{l}{g}}$

Here $\omega$ is angular frequency so, T = 2π/$\omega$

Hence, $T>2\pi\sqrt{\frac{l}{g}}$

so if we do not use the approximation the result for time period would not be same and actual time period is greater than calculated one, because the formula for time period, T = 2π√{l/g} is valid when angle of oscillation is very small.

Answer 2

Answer:

(a) given, mass of the automobile, M = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

So, equation for the restoring force for the system is given by, F = –4kx = mg

Where, k is the spring constant of the suspension system

so, equivalent spring constant = 4k

Time period, T = 2π √{m/4k}

and k = mg/4x

= (3000 × 10)/ (4 × 0.15)

= 5000 = 5 × 10⁴ Nm

Hence, spring Constant, k = 5 × 10⁴ Nm

(b) Each wheel supports a mass, m = 3000/4 = 750 kg

For damping factor b, the equation for displacement is written by x=x_0e^{-\frac{bt}{2m}}x=x

0

e

−

2m

bt

The amplitude of oscilliation decreases by 50 %.

∴ x = 50 % of x_0x

0

= x_0x

0

/2

x_0/2=x_0e^{-\frac{bt}{2m}}x

0

/2=x

0

e

−

2m

bt

ln2 = bt/2m

∴ b = 2m ln2/t where, t is time period.

e.g., t = 2π√{m/4k} = 2 × 3.14 √{3000/(4 × 5 × 10⁴)} = 0.7691s