A mass m is suspended from a spring- Its frequency of oscillation is f- The spring is into equal halves and the same mass is suspended from one of the two pieces of the spring- The frequency of oscillation is the mass will besqrt-2- f2ff-2f

Question

Toppr Admin · Accepted Answer

-Let initial-spring constant be k- Then- we can say-f-12-x3C0-x221A-kmAfter the spring is cut in two halves-spring constant becomes 2k- Then-f2-12-x3C0-x221A-2km-x221A-2f- Option A is thus correct-

A mass $m$ is suspended from a spring. Its frequency of oscillation is $f$ . The spring is cut into equal halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation is the mass will be
$\sqrt{2} f$
$2 f$
$f / 2$
$f$

Let initial spring constant be k. Then, we can say,
$f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$
After the spring is cut in two halves, spring constant becomes 2k. Then,
$f_{2} = \frac{1}{2 π} \sqrt{\frac{2 k}{m}} = \sqrt{2} f$ . Option A is thus correct.

A mass m is suspended from a spring. Its frequency of oscillation is f. The spring is cut into equal halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation is the mass will be√2f2ff/2f

Let initial spring constant be k. Then, we can say,f=12π√kmAfter the spring is cut in two halves, spring constant becomes 2k. Then,f2=12π√2km=√2f. Option A is thus correct.

A mass $m$ is suspended from a spring. Its frequency of oscillation is $f$ . The spring is cut into equal halves and the same mass is suspended from one of the two pieces of the spring. The frequency of oscillation is the mass will be
$\sqrt{2} f$
$2 f$
$f / 2$
$f$

Let initial spring constant be k. Then, we can say,
$f = \frac{1}{2 π} \sqrt{\frac{k}{m}}$
After the spring is cut in two halves, spring constant becomes 2k. Then,
$f_{2} = \frac{1}{2 π} \sqrt{\frac{2 k}{m}} = \sqrt{2} f$ . Option A is thus correct.