How do you find what the mass on the spring is if you know the period and force constant of the harmonic oscillator?

Damon Vazquez
2022-09-01
Answered

You can still ask an expert for help

Cameron Wallace

Answered 2022-09-02
Author has **8** answers

Start with the equation for the period

$T=2\pi \sqrt{\frac{m}{k}}$, where

T-the period of oscillation;

m- the mass of the oscillating object;

k- a constant of proportionality for a mass on a spring;

You need to solve this equation for m, so start by squaring both sides of the equation

${T}^{2}=(2\pi \ast \sqrt{\frac{m}{k}}{)}^{2}$

${T}^{2}=(2\pi {)}^{2}\ast (\sqrt{\frac{m}{k}}{)}^{2}$

${T}^{2}=4{\pi}^{2}\ast \frac{m}{k}$

Now all you have to do is isolate m on one side of the equation

${T}^{2}\ast k=4{\pi}^{2}\ast m$

$m=\frac{{T}^{2}\ast k}{4{\pi}^{2}}=k\ast \frac{{T}^{2}}{4{\pi}^{2}}$

$T=2\pi \sqrt{\frac{m}{k}}$, where

T-the period of oscillation;

m- the mass of the oscillating object;

k- a constant of proportionality for a mass on a spring;

You need to solve this equation for m, so start by squaring both sides of the equation

${T}^{2}=(2\pi \ast \sqrt{\frac{m}{k}}{)}^{2}$

${T}^{2}=(2\pi {)}^{2}\ast (\sqrt{\frac{m}{k}}{)}^{2}$

${T}^{2}=4{\pi}^{2}\ast \frac{m}{k}$

Now all you have to do is isolate m on one side of the equation

${T}^{2}\ast k=4{\pi}^{2}\ast m$

$m=\frac{{T}^{2}\ast k}{4{\pi}^{2}}=k\ast \frac{{T}^{2}}{4{\pi}^{2}}$

aurelegena

Answered 2022-09-03
Author has **2** answers

Let's say we started from $\omega =\sqrt{\frac{k}{m}}$. It's a bit different but a similar approach.

If we examine the equation

$y=A\mathrm{sin}(n\theta +\varphi )+k$

If n was doubled, the frequency would be doubled, but the period would be halved. So, we know that $\omega \propto \frac{1}{T}$

If $\omega $ is $2\pi rad/s$, the period T is 1 s, so to create the equality between the two variables, we match up the units by multiplying $\frac{1}{T}$ by $2\pi rad$ to get $\omega =\frac{2\pi}{T}$

$\omega =\frac{2\pi}{T}=\sqrt{\frac{k}{m}}$

Square both sides:

$\frac{4{\pi}^{2}}{{T}^{2}}=\frac{k}{m}$

Reciprocate both sides and then multiply by k:

$m=\frac{k{T}^{2}}{4{\pi}^{2}}$

If we examine the equation

$y=A\mathrm{sin}(n\theta +\varphi )+k$

If n was doubled, the frequency would be doubled, but the period would be halved. So, we know that $\omega \propto \frac{1}{T}$

If $\omega $ is $2\pi rad/s$, the period T is 1 s, so to create the equality between the two variables, we match up the units by multiplying $\frac{1}{T}$ by $2\pi rad$ to get $\omega =\frac{2\pi}{T}$

$\omega =\frac{2\pi}{T}=\sqrt{\frac{k}{m}}$

Square both sides:

$\frac{4{\pi}^{2}}{{T}^{2}}=\frac{k}{m}$

Reciprocate both sides and then multiply by k:

$m=\frac{k{T}^{2}}{4{\pi}^{2}}$

asked 2022-11-18

A string is made to vibrate at its third harmonic. The diagram shows two points P and Q at a particular instant in time. Which of the following compares the period of vibration of P and Q?

(The string is closed at both ends; P is just before the first maximum antinode and Q is at the next minimum antinode).

A. Period of P and Q is the same

B. Period of P and Q is different

Why is it A not B?

(The string is closed at both ends; P is just before the first maximum antinode and Q is at the next minimum antinode).

A. Period of P and Q is the same

B. Period of P and Q is different

Why is it A not B?

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