Question

A mass m attached to a spring is subjected to a harmonic force as shown in figure. The amplitude of the forced motion is observed to be 50 mm. The value of m (in kg) is

• A. 0.1
• B. 1.0
• C. 0.3
• D. 0.5

Answer 1

The correct option is A 0.1

Method I:
Given,
$F\left(t\right)=100cos\left(100t\right)N$

$F_0=100N$

$\omega =100rad/s$

$k=3000N/m$

Amplitude :
$X=50\times {10}^{-3}m$

$\text{Now by}X=\frac{F_0}{(k-m{\omega }^2)}$

As damping is zero.

$k-m{\omega }^2=\frac{100}{50\times {10}^{-3}}=2000$

$m{\omega }^2=3000-2000$

$m=\frac{1000}{(100{)}^2}=0.1kg$

Method II :
Amplitude,

$A=\frac{\frac{F_0}{k}}{\sqrt{{[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2]}^2+{\left(\frac{2\xi \omega }{{\omega }_n}\right)}^2}}$

For no damping,

$\xi =0$
$A=\frac{\frac{F_0}{k}}{\sqrt{{[1-{\left(\frac{\omega }{{\omega }_n}\right)}^2]}^2}}=\frac{\frac{F_0}{k}}{1-{\left(\frac{\omega }{{\omega }_n}\right)}^2}$

$0.05=\frac{\frac{100}{300}}{1-{\left(\frac{100}{{\omega }_n}\right)}^2}$

$\therefore$ ${\omega }_n=100\sqrt{3}$

But, we know that,

${\omega }_n=\sqrt{\frac{k}{m}}$

$\therefore m=\frac{k}{{\omega }_n^2}$

$=\frac{3000}{3\times 100\times 100}=0.1kg$