A second\'s pendulum is a simple pendulum whose period is 2 seconds: hence. one

Question

A second's pendulum is a simple pendulum whose period is 2 seconds: hence. one which will beat out seconds as it swings from side to side. Using the equation for the acceleration due to gravity, calculate the length of such a pendulum if g = 980 cm/sec^2. Can you think of any reasons why this pendulum would not the a good standard for the unit of length, i.e.. measuring its string when its period is exactly 2 seconds? Explain with justification using physics or math principles why the angle through which the pendulum swings must be no more than about 30^degree. Explain physically why it was unnecessary to measure the mass of the sphere for any case (any mass).

Explanation / Answer

given

T = 2 s

g = 980 cm/s^2

L = ?

we know, Time period of a pendulum, T = 2*pi*sqrt(L/g)

T^2 = 4*pi^2*L/g

==> L = g*T^2/(4*pi^2)

= 980*2^2/(4*pi^2)

= 99.3 cm

This pendulum is not a good standard. because, g values is not a constant through out the earth surface.

when we derive equation, T = 2*pi*sqrt(L/g)

we take small angle approximation.

when theta is small, sin(theta) = theta

In the eqation, T = 2*pi*sqrt(L/g)

mass does not appear. It means the period of a pendulum does not depend on length of the pendulum.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.