Question

two shms of masses 10 gram and 899 gram oscillate separately under the action of same restoring force.calculate their ratio of their frequencies

Answer 1

The ratio of frequencies for two objects in simple harmonic motion with masses of 10 grams and 899 grams, when subjected to the same restoring force, can be calculated using the inverse square root of the ratio of their masses.

Step-by-step explanation:

The frequency of simple harmonic motion (SHM) for a mass attached to a spring is given by the formula f = \( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where f is the frequency, k is the spring constant, and m is the mass of the object. Since the restoring force (spring constant k) is the same for both objects, the ratio of their frequencies can be determined by the inverse square root of the ratio of their masses. Thus, for masses 10 grams and 899 grams, the ratio of their frequencies will be f1/f2 = \( \sqrt{\frac{m2}{m1}} \) = \( \sqrt{\frac{899}{10}} \), which simplifies to the final ratio of their frequencies. It’s important to keep in mind that higher mass results in a lower frequency when the spring constant is held constant.

Answer 2

The ratio of frequencies of two simple harmonic motions (SHMs) with masses 10g and 899g can be calculated using the formula f1/f2 = √(m2/m1). The ratio of their frequencies is approximately 9.48.

Step-by-step explanation:

The ratio of frequencies of two simple harmonic motions (SHMs) can be calculated using the formula:

f1/f2 = √(m2/m1)

Where f1 and f2 are the frequencies of the two SHMs, and m1 and m2 are the masses of the respective oscillators.

Applying the formula to the given problem:

f1/f2 = √(m2/m1)

f1/f2 = √(899g/10g) = √(899/10) = √89.9 ≈ 9.48

Therefore, the ratio of the frequencies of the two SHMs is approximately 9.48.