Question

A small pan suspended from a helical spring as shown in the figure, empty has an observed period of vibration $T_0$. When carrying a known weight P, the observed period of vibration is $T_1$ and when carrying an unknown weight W (without P), the observed period is $T_2$. Find the magnitude of the weight W.

• A. $\left(\frac{T_2+T_1}{2T_0}\right)P$
• B. $\left(\frac{T_2+T_1}{T_1+T_0}\right)P$
• C. $\left(\frac{T_2+T_1}{T_1+T_0}\right)P$
• D. $\left(\frac{T_2^2+T_0^2}{T_1^2+T_0^2}\right)P$

Answer 1

The correct option is D $\left(\frac{T_2^2+T_0^2}{T_1^2+T_0^2}\right)P$

Let the weight of the pan be Q
Then the period of oscillations



$T_0=2\pi \sqrt{\frac{Q}{kg}}$

$(T_0{)}^2=\frac{4{\pi }^2}{kg}Q.....(1)$

When the known weight P is placed on the pan

$T_1=2\pi \sqrt{\frac{P+Q}{kg}}$

$T_1^2=\frac{4{\pi }^2}{kg}(P+Q)......\left(2\right)$

$\text{Using Eq.(1),}{T}_1^2=\frac{4{\pi }^2}{kg}{P}^2+T_0$

$T_1^2-T_0^2=\frac{4{\pi }^2}{kg}P.....\left(3\right)$

When the unknown weight W is placed on the pan

$T_2=2\pi \sqrt{\frac{W+Q}{kg}}$

$T_2^2=\frac{4{\pi }^2}{kg}(W+Q)=\frac{4{\pi }^2}{kg}W+T_0^2$

$T_2^2-T_0^2=\frac{4{\pi }^2}{kg}W....\left(4\right)$

Dividing Eq. (4) by Eq.(3)

$\frac{T_2^2-T_0^2}{T_1^2-T_0^2}=\frac{W}{P}$

$\therefore \text{The unknown weight,}$
$W=\left(\frac{T_2^2+T_0^2}{T_1^2+T_0^2}\right)P$