A particle at the end of a spring executes si | vedclass

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

Question

(B) The period of a mass $m$ attached to a spring of constant $k$ is given by $t = 2\pi \sqrt{\frac{m}{k}}$.Thus,$t_1 = 2\pi \sqrt{\frac{m}{k_1}}$ and

Vedclass · Accepted Answer

$T^2 = t_1^2 + t_2^2$

Vedclass · Answer

(B) The period of a mass $m$ attached to a spring of constant $k$ is given by $t = 2\pi \sqrt{\frac{m}{k}}$.Thus,$t_1 = 2\pi \sqrt{\frac{m}{k_1}}$ and $t_2 = 2\pi \sqrt{\frac{m}{k_2}}$.Squaring these,we get $t_1^2 = 4\pi^2 \frac{m}{k_1}$ and $t_2^2 = 4\pi^2 \frac{m}{k_2}$.When two springs are connected in series,the equivalent spring constant $k_{eq}$ is given by $\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}$.The period of oscillation for the series combination is $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.Squaring this,$T^2 = 4\pi^2 \frac{m}{k_{eq}} = 4\pi^2 m \left( \frac{1}{k_1} + \frac{1}{k_2} \right)$.Substituting the expressions for $t_1^2$ and $t_2^2$,we get $T^2 = 4\pi^2 \frac{m}{k_1} + 4\pi^2 \frac{m}{k_2} = t_1^2 + t_2^2$.

$A$ particle at the end of a spring executes simple harmonic motion with a period $t_1$,while the corresponding period for another spring is $t_2$. If the period of oscillation with the two springs in series is $T$,then

Similar Questions

$2 \ mol$ of zinc is dissolved in $HCl$ at $25 \ ^\circ C$. The work done in an open vessel is:

Electrons with a kinetic energy of $6.023 \times 10^4 \ J/mol$ are evolved from the surface of a metal when it is exposed to radiation of wavelength $600 \ nm$. The minimum amount of energy required to remove an electron from the metal atom is

If $\int \frac{d x}{x(\log x-2)(\log x-3)}=I+C$,then $I$ is equal to

Which of the following has the highest density?

Vedclass Test Series

Exam Paper Generator

Online Exam Module