Question

A sonometer wire under a tension of 40 N , vibrates in unison with a tuning fork of frequency 384 Hz. find the number of beats produced in 2 sec when the tension in the wire is reduced by 1.24 N . 2.] A conical pendulum has length 1 m and bob of mass 0.1 kg the angular speed of the bob is 14/√10 rad/sec, find the tension in the string.

Answer 1

Answer on Question #47014-Physics-Mechanics-Kinematics-Dynamics

A sonometer wire under a tension of $T = 40 \, \text{N}$, vibrates in unison with a tuning fork of frequency $N = 384 \, \text{Hz}$. Find the number of beats produced in 2 sec when the tension in the wire is reduced by $1.24 \, \text{N}$.

Solution

Let N is a frequency of tuning fork. Then frequency of wire when a tension is $40 \, \text{N}$ is N.

The frequency of wire when a tension is reduced by $1.24 \, \text{N}$ is N-k. Hence,

$$N = \frac{1}{2L} \sqrt{\frac{T}{m}}; \quad N - k = \frac{1}{2L} \sqrt{\frac{T - \Delta T}{m}}.$$ $$\frac{N - k}{N} = \sqrt{\frac{T - \Delta T}{T}} \rightarrow k = N \left(1 - \sqrt{\frac{T - \Delta T}{T}}\right) = 384 \left(1 - \sqrt{\frac{40 - 1.24}{40}}\right) = 6 \frac{\text{beats}}{s} = 12 \frac{\text{beats}}{2s}.$$

Answer: 12.

2.] A conical pendulum has length 1 m and bob of mass 0.1 kg the angular speed of the bob is 14/V10 rad/sec, find the tension in the string.

Solution

The object is subject to two forces: the gravitational force $mg$ which acts vertically downwards, and the tension force $T$ which acts upwards along the string. The tension force can be resolved into a component $T \cos\theta$ which acts vertically upwards, and a component $T \sin\theta$ which acts towards the centre of the circle. Force balance in the vertical direction yields

$$T \cos\theta = mg.$$

Since the object is executing a circular orbit, radius $r$, with angular velocity $\omega$, it experiences a centripetal acceleration $\omega^2 r$. Hence, it is subject to a centripetal force $m\omega^2 r$. This force is provided by the component of the string tension which acts towards the centre of the circle. In other words,

$$T \sin\theta = m\omega^2 r.$$

Note that if $l$ is the length of the string then $r = l \sin\theta$. It follows that

$$T = m\omega^2 l = 0.1 \cdot \left(\frac{14}{\sqrt{10}}\right)^2 \cdot 1 = 1.96 \, \text{N}.$$

Answer: 1.96 N.

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