## Question

### Solution

Correct option is

200 Hz

The ends of the string x = 0 and x = 100 cm are nodes. The string is plucked at x = 25 cm. Hence x = 25 cm is an antinode. The next antinode is at x = 75 cm. Therefore, the string vibrates in 2 segments. Hence its frequency is

#### SIMILAR QUESTIONS

Q1

A spring is extended by a length l, the according to Hooke’s law

Q2

A transverse wave is represented by

For what value of λ is the maximum particle velocity equal to twice the wave velocity?

Q3

A pendulum clock keeps correct time at 20oC. The coefficient of linear expansion of pendulum is . If the room temperature increases to 40oC, how many seconds will the clock lose or gain per day?

Q4

A mass m is suspended from a spring of negligible mass. The system oscillates with frequency n. What will be the frequency if a mass 4 m is suspended from the same spring?

Q5

In a sinusoidal wave, the time required for a particular point to move from maximum displacement to zero displacement is 0.17s. The frequency of the wave is

Q6

The kinetic energy of a particle executing S.H.M. is 16 J, when it is at its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the time period of oscillation is

Q7

A standing wave having 3 nodes and 2 antinodes is formed between two atoms having a separation of 1.21 Å between them. The wavelength of the standing wave is

Q8

A body is executing S.H.M. with angular frequency 2 rad s–1. If the amplitude of the motion is 60 mm, the velocity of the body at 20 mm displacement is

Q9

A wave is represented by the equation

Where x is in metres and t in seconds. The expression represents

Q10

Two bodies M and N of equal masses are suspended from two separate springs of force constants k1 and k2 respectively. If they oscillate with equal maximum velocities, the amplitudes of M and N will be the ratio of