Question

A mass on a string of unknown length oscillates as a pendulum with a period of $4.0s$.

What is the period if

a. The mass is doubled?

b. The string length is doubled?

c. The string length is halved?

d. The amplitude is doubled?

Short answer

The period if the mass is doubled$T=4.00\mathrm{s}$.

The period if the string is doubled$T=5.66\mathrm{s}$.

The period if the string length is halved$T=2.83\mathrm{s}$.

The period if the amplitude is doubled$T=4.00\mathrm{s}$.

Step-by-step solution

 Principles.

The period $T$of an easy pendulum of length $L$is given by,

$T=2\pi \sqrt{\frac{L}{g}}$

where $g=9.80\mathrm{m}/{\mathrm{s}}^2$is that the acceleration of gravity.

Note that the amount of a straightforward pendulum depends only on its length and the magnitude of its constant of gravitation.

It doesn't rely upon the mass of the item hanging from the sting or the amplitude of vibration.

The given data.

The original period of the pendulum is: $T_0=4.00\mathrm{s}$.

 Required Data.

In part (a), we are asked to determine the period of the pendulum if the mass is doubled.

In part (b), we are asked to determine the period of the pendulum if the string length is doubled.

In part (c), we are asked to determine the period of the pendulum if the string length is halved.

In part (d), we are asked to determine the period of the pendulum if the amplitude is halved.

Solution.

(a) The period of the pendulum is independent of the mass, therefore, the period when the mass is doubled is

$T=T_0=4.00\mathrm{s}$

(b) Let $L$be the new length of the pendulum and $L_0$be the original length of the pendulum. The period of the pendulum if the string length is doubled is found by substituting $2L_0$for $L$in Equation $(*)$:

$T=2\pi \sqrt{\frac{2L_0}{g}}$

$=\sqrt{2}\left(2\pi \sqrt{\frac{L_0}{g}}\right)$

where the term in parenthesis is the original period of the pendulum $T_0$:

$T=\sqrt{2}{T}_0$

$=\sqrt{2}(4.00\mathrm{s})$

$=5.66\mathrm{s}$

The period of the pendulum if the string length is halved is found by substituting $L_0/2$for $L$in Equation $(*)$:

$T=2\pi \sqrt{\frac{L_0/2}{g}}$

$=\frac{1}{\sqrt{2}}\left(2\pi \sqrt{\frac{L_0}{g}}\right)$

where the term in parenthesis is the original period of the pendulum $T_0$:

$T=\frac{T_0}{\sqrt{2}}$

$=\frac{4.00\mathrm{s}}{\sqrt{2}}$

$=2.83s$

The period of the pendulum is independent of the amplitude, therefore, the period when the amplitude is halved is

$T=T_0=4.00\mathrm{s}$