A clock S is based on the oscillation of a sp | vedclass

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

Question

(B) The frequency of the spring-based clock $S$ is given by $f_S = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$. This frequency depends only on the spring const

Vedclass · Accepted Answer

$P$ will run faster than $S$

Vedclass · Answer

(B) The frequency of the spring-based clock $S$ is given by $f_S = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$. This frequency depends only on the spring constant $k$ and mass $m$,and is independent of the acceleration due to gravity $g$.The frequency of the pendulum-based clock $P$ is given by $f_P = \frac{1}{2\pi} \sqrt{\frac{g}{l}}$. This frequency depends on the acceleration due to gravity $g$.The acceleration due to gravity is $g = \frac{GM}{R^2} = \frac{G}{R^2} \cdot \rho \cdot \frac{4}{3} \pi R^3 = \frac{4}{3} \pi \rho G R$.Since the density $\rho$ remains the same and the radius $R$ is doubled,the new acceleration due to gravity $g'$ becomes $2g$.Consequently,the frequency of the pendulum clock $P$ becomes $f_P' = \sqrt{2} f_P$,while the frequency of the spring clock $S$ remains unchanged.Therefore,the pendulum clock $P$ will run faster than the spring clock $S$.

Column $I$	Column $II$
$(A)$ Potential energy of a simple pendulum ($y$-axis) as a function of displacement ($x$-axis)	$(p)$ Parabolic curve opening upwards
$(B)$ Displacement ($y$-axis) as a function of time ($x$-axis) for a one-dimensional motion at zero or constant acceleration	$(q)$ Linear graph passing through origin
$(C)$ Range of a projectile ($y$-axis) as a function of its velocity ($x$-axis) when projected at a fixed angle	$(r)$ Linear graph with non-zero intercept
$(D)$ The square of the time period ($y$-axis) of a simple pendulum as a function of its length ($x$-axis)	$(s)$ Parabolic curve opening upwards (starting from origin)

$A$ clock $S$ is based on the oscillation of a spring,and a clock $P$ is based on pendulum motion. Both clocks run at the same rate on Earth. On a planet having the same density as Earth but twice the radius,which of the following is true?

Similar Questions

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

$A$ body executing simple harmonic motion has a maximum acceleration equal to $24 \, m/s^2$ and a maximum velocity equal to $16 \, m/s$. The amplitude of the simple harmonic motion is:

$A$ mass $m$ oscillates with simple harmonic motion with frequency $f = \frac{\omega}{2\pi}$ and amplitude $A$ on a spring with constant $K$. Therefore:

Vedclass Test Series

Exam Paper Generator

Online Exam Module