Question

ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is at rest as it hangs from the string. When the object is raised to a height above its lowest point and released from rest, the object undergoes simple harmonic motion with a frequency f0. In a second scenario, the length of the string is cut in half before the object undergoes simple harmonic motion again. What is the new frequency of oscillation of the object in terms of f0?

Answer 1

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

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

where T is time period , l is length of pendulum and g is acceleration due to gravity .

$\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }$  , f is frequency of oscillation

For the given case

$\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }$

subsequently length becomes half so

$\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }$

dividing

$\frac{f}{f_o} = \sqrt{\frac{2}{1} }$

f = $\sqrt{2} f_o$

frequency of oscillation becomes √2 times.