A body executes simple harmonic motion under | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$,the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = ma = -kx$,the force con

Vedclass · Accepted Answer

$\frac{12}{25}$

Vedclass · Answer

(A) For a body of mass $m$ executing $SHM$ under a force $F = -kx$,the time period is $T = 2\pi \sqrt{\frac{m}{k}}$.Since $F = ma = -kx$,the force constant $k$ is proportional to $\frac{1}{T^2}$ (as $k = m\omega^2 = m(\frac{2\pi}{T})^2$).Thus,$k_1 \propto \frac{1}{T_1^2}$ and $k_2 \propto \frac{1}{T_2^2}$.When both forces act in the same direction,the effective force constant is $k_{eff} = k_1 + k_2$.The new time period $T$ is given by $\frac{1}{T^2} = \frac{1}{T_1^2} + \frac{1}{T_2^2}$.Substituting the values: $\frac{1}{T^2} = (\frac{5}{4})^2 + (\frac{5}{3})^2 = \frac{25}{16} + \frac{25}{9}$.$\frac{1}{T^2} = 25 	imes (\frac{9+16}{144}) = 25 	imes \frac{25}{144} = \frac{625}{144}$.Taking the square root: $\frac{1}{T} = \frac{25}{12}$,which gives $T = \frac{12}{25} \ s$.

Similar Questions

If $A$ and $B$ are two events such that $A \subset B$ and $P(B) \neq 0$,then which of the following is correct?

The $IUPAC$ name of the given compound is:

$A$ body weighs $72 \ N$ on the surface of the earth. What is the gravitational force (in $N$) on it,at a height equal to half the radius of the earth?

$A$ circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module